Chapter 3.22 - Solved Examples Related to Sine Rule and Cosine Rule

In the previous section, we completed a discussion on cosine formula. In this section we will see Napiers' Analogies.

• The list of trigonometric identities can be seen here.

◼ In any triangle, the following three relations are applicable:
$\begin{eqnarray}
&\text{(i)}\;\tan \frac{B-C}{2}&=\frac{b-c}{b+c}\, \cot \frac{A}{2} \\
&\text{(ii)}\;\tan \frac{C-A}{2}&=\frac{c-a}{c+a}\, \cot \frac{B}{2} \\
&\text{(iii)}\;\tan \frac{A-B}{2}&=\frac{a-b}{a+b}\, \cot \frac{C}{2} \
\end{eqnarray}$

Proof for (i) can be written in 3 steps:
1. Consider the three ratios in the sin formula.
• Each of those three ratios will give the same value for a triangle under consideration.
• So we can say that, any triangle will have it's own unique constant value 'k' which will be equal to each of the ratio in the sine formula.
• We can write: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
• From this we get: a = k sin A, b = k sin B, c = k sin C
2. Consider the first analogy. $\frac{b-c}{b+c}$ can be modified as:
$\begin{eqnarray}
&\frac{b-c}{b+c}& =\frac{k \sin B-k \sin C}{k \sin B+k \sin C} \\
&{}& =\frac{k (\sin B- \sin C)}{k(\sin B+ \sin C)}\;=\frac{\sin B- \sin C}{\sin B+ \sin C} \\
&{}& =\frac{2\cos \frac{B+C}{2} \sin \frac{B-C}{2}}{2\sin \frac{B+C}{2} \cos \frac{B-C}{2}} \\
&{}& \text{(Using identity 20.d)} \\
&{}& =\cot \frac{B+C}{2} \tan \frac{B-C}{2} \\
&{}& =\cot \left(\frac{\pi}{2}-\frac{A}{2} \right) \tan \left(\frac{B-C}{2} \right) \\
&{}& \left[ \because \frac{\pi}{2}=\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\right] \\
&{}& =\tan \left(\frac{A}{2} \right) \tan \left(\frac{B-C}{2} \right) \\
&{}& \left[ \because \text{using identities 5 and 6, }\cot \left(\frac{\pi}{2}-\frac{A}{2} \right)=\tan \left(\frac{A}{2} \right)\right] \\
&{}& =\frac{\tan \frac{B-C}{2} }{\cot \frac{A}{2}} \
\end{eqnarray}$
3. Thus we get: $\tan \frac{B-C}{2}=\frac{b-c}{b+c}\, \cot \frac{A}{2}$
• In the same way, we can prove (ii) and (iii) also.

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Let us see the application of the above analogies. It can be written in steps:
1. Suppose that, we are given two sides b and c, and the included angle A
• Then we can easily calculate the RHS of (i)
2. That means, we can easily obtain $\tan \frac{B-C}{2}$
• From that, we can obtain $\frac{B-C}{2}$
• From that, we can obtain (B-C)
3. We are already given A.
• Using that A, we can find (B+C)
   ♦ Because, (B+C) = 180 - A
4. Thus we have two equations:
   ♦ One involving B+C
   ♦ The other involving B-C
• Using those equations, we can calculate B and C

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Let us see some solved examples:

Solved example 3.97
In any triangle ABC, prove that
$a \sin(B-C)+b \sin(C-A)+c \sin(A-B)=0$
Solution:
1. We have: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
• From this we get: a = k sin A, b = k sin B, c = k sin C
2. So we can modify the LHS of the given equation as:
k sin A sin(B-C) + k sin B sin(C-A) + k sin C sin(A-B)
3. Now we apply identity 8 to each of the three terms. We get:
k sin A [sin B cos C - cos B sin C] + k sin B [sin C cos A - cos c sin A] + k sin C [sin A cos B - cos A sin B]
4. Expanding this, we get:
k sin A sin B cos C - k sin A cos B sin C + k sin B sin C cos A - k sin B cos C sin A + k sin C sin A cos B - k sin C cos A sin B
• This is same as:
k [sin A sin B cos C - sin A cos B sin C + sin B sin C cos A - sin B cos C sin A + sin C sin A cos B - sin C cos A sin B]
5. There are identical terms in the above expansion. The first set is underlined below:
k [sin A sin B cos C - sin A cos B sin C + sin B sin C cos A - sin B cos C sin A + sin C sin A cos B - sin C cos A sin B]
• Being opposite in signs, they will cancel each other.
• The second set is under lined below:
k [- sin A cos B sin C + sin B sin C cos A + sin C sin A cos B - sin C cos A sin B]
• Being opposite in signs, they will cancel each other.
• The remaining set is:
k [+ sin B sin C cos A - sin C cos A sin B]
• Clearly, they will also cancel each other
6. In effect, the LHS becomes: k[0] = 0
• Thus we get LHS = RHS

Solved example 3.98
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45^o and from a point B, the angle of elevation is 60^o, where B is a point at a distance of d from the point A measured along the line AB which makes an angle 30^o with AQ. Prove that d = h(√3-1)
Solution:
1. First, the instrument (used for measuring angles) is placed at A.
• When measured from A, the angle is 45^o. This is shown in fig.3.45(i) below:

Fig.3.45

2. Then the instrument is taken to a point B.
• The distance of B from A is d. Also, AB makes an angle of 30^o with PQ.
• When measured from B, the angle is 60^o.
• These details must be added to the first fig. The modified fig. is fig.3.45(ii) above. 
3. Let us calculate some important angles and sides:
(i) In  ◺APQ, we have: ∠APQ = (180 - 90 - 45) = 45^o
So it is an isosceles right triangle. We get: AQ = PQ = h
(ii) Since APQ is a right triangle, we get: $AP=\sqrt{AQ^2+PQ^2}=\sqrt{h^2+h^2}=\sqrt{2h^2}=\sqrt{2}\,h$
(iii) Inside  ◺APQ, we have: ∠PAB = (45 - 30) = 15^o
(iv) In ◺ PBC, we have: ∠BPC = (180 - 90 - 60) = 30^o
(v) Inside ◺APQ,  ∠APB = (45 - 30) = 15^o
(vi) From (ii) and (iv), it is clear that    ⃤⃤  APB is an isosceles triangle.
We have: AB = PB = d
(vii) Also in    ⃤⃤  APB, we get: ABP = (180 - 15 -15) = 150^o
4. So from 3(ii), (iii), (v), (vi) and (vii), we have all the angles and sides of    ⃤⃤  APB.
This is shown in fig.3.45(iii).
• Now we can apply sine rule. We get:
$\frac{AP}{\sin B}=\frac{BP}{\sin A} \Rightarrow \frac{\sqrt{2}\,h}{\sin 150}=\frac{d}{\sin 15} \Rightarrow d=\frac{\sqrt{2}\,h \times \sin 15}{\sin 150}$
5. We need sin 15 and sin 150:
• We know that: sin 150 = sin (180 - 150) = sin 30 = $\frac{1}{2}$
• We have calculated sin 15 in an earlier section. (Solved example 3.30 in section 3.13)
   ♦ We got: $\sin 15 = \frac{\sqrt{3} - 1}{2 \sqrt 2}$
6. So the result in (4) becomes:
$\begin{eqnarray}
&{}& d=\frac{\sqrt{2}\,h \times \sin 15}{\sin 150} \\
&\Rightarrow& d=\frac{\sqrt{2}\,h \times \frac{\sqrt{3} - 1}{2 \sqrt 2}}{\frac{1}{2}} \\
&\Rightarrow& d=\frac{h \times \frac{\sqrt{3} - 1}{2}}{\frac{1}{2}} \\
&\Rightarrow& d=h(\sqrt{3} - 1) \
\end{eqnarray}$

Solved example 3.99
A lamp post is situated at the middle point M of the side AC of a triangular plot ABC with BC = 7 m, CA = 8 m and AB = 9 m. Lamp post subtends an angle 15^o at the point B. Determine the height of the lamp post.
Solution:
1. In fig.3.46(i), the triangular plot ABC is lying on the ground. The plot is shown in red color.

Fig.3.46

2. Mark M, the midpoint of AC. Erect a perpendicular PM at M.
• PM is the post. This is shown in fig.(ii). Given that: ∠PBM = 15^o
• We are asked to find the height PM.
3. First we need to find ∠BAM
• This can be easily calculated by solving    ⃤⃤ ABC. Since only the sides of     ⃤⃤ ABC are given, we need to apply cosine rule. We get:
BC2` = AB2` + AC2` - 2 AB.AC cos (∠BAC)
• Substituting the known values, we get:
72` = 92` + 82` - 2 × 9 × 8 × cos (∠BAC)
• Thus we get: cos (∠BAC) = cos A = 0.6667
    ♦ Then A = 48.19^o
• Consider the identity: cos x = cos (360-x)
    ♦ We get: cos 48.19 = cos (360-48.19) = cos 311.81 = 0.6667
    ♦ That means, A can be 48.19 or 311.81
    ♦ 311.81 is not acceptable because, it is greater than 180
• Thus we get: A = 48.19^o
4. Now consider    ⃤⃤ BAM.
• In this triangle, we have the measurements of two sides AB and AM, and the included angle A.
• So we can apply the cosine rule and find BM.
• We get: BM2` = AB2` + AM2` - 2 AB × AM × cos A
• Substituting the known values, we get:
BM2` = 92` + 42` - 2 × 9 × 8 × cos 48.19
• Thus we get: BM = 7 m
5. Finally, consider    ⃤⃤ PBM
• We have: tan (∠PBM) = $\frac{PM}{BM}=\frac{PM}{BM}$
• Given that: ∠PBM = 15^o
• Thus we get: tan 15 = $\frac{PM}{7}$
6. We have calculated tan 15 earlier as: $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• We can write: $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{PM}{7}$
• Thus we get: PM = $\frac{7(\sqrt{3}-1)}{\sqrt{3}+1}$
⇒ PM = $\frac{7(\sqrt{3}-1)}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
⇒ PM = $\frac{7(\sqrt{3}-1)^2}{2}$
⇒ PM = $\frac{7(3+1-2 \sqrt{3})}{2}=\frac{7(4-2 \sqrt{3})}{2}$
⇒ PM = $\frac{7\times 2(2- \sqrt{3})}{2}=7(2- \sqrt{3})$ m

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Link to some more solved examples is given below:

Solved example 3.100 to 3.114

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We have completed the discussion in this chapter. In the next chapter, we will see mathematical induction.

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Chapter 3.21 -The Cosine Formula

In the previous section, we completed a discussion on sine formula. We saw that, we will not be able to apply the sine rule when:
(i) We are given two sides and the angle included between those two given sides.
(ii) We are given all the three sides but no angles.
• In such situations, we can use the cosine formula.

• We can use the same notations and figures 3.39, 40 and 41 that we saw in the previous section. So we will show them again.

• Consider the triangle ABC in fig.3.39 below.
• For this triangle, we can define three angles:
   ♦ Angle A is the angle between sides AB and AC 
   ♦ Angle B is the angle between sides BA and BC 
   ♦ Angle C is the angle between sides CA and CB 
• We can define three sides also:
   ♦ a is the length of the side BC (The side opposite angle A)
   ♦ b is the length of the side AC (The side opposite angle B)
   ♦ c is the length of the side AB (The side opposite angle C)

Fig.3.39

◼ Based on the above notations, we can write the cosine formula as:

a² = b² + c² - 2bc cos A
b² = c² + a² - 2ca cos B
c² = a² + b² - 2ab cos C

• For proving this formula, we will analyze an obtuse triangle and an acute triangle.

• An obtuse triangle is that triangle in which one angle is obtuse and the remaining two are acute. (Remember that, no triangle can have more than one obtuse angles). • An acute triangle is that triangle in which all three angles are acute.

• Both triangles can be analyzed using the same steps.
• When the formula is proved for both those types of triangles, we can confidently apply it to any triangle.

The proof can be written in 8 steps:
1. Dropping perpendiculars from the top vertex:
• For the obtuse triangle in fig.3.40(i) below, the perpendicular dropped from the top vertex B can meet the base AC only if we extend the base a bit towards the right.

Fig.3.40

• For the acute triangle in fig.3.40(ii), the perpendicular dropped from the top vertex B can meet the base AC with out any difficulty.
2. Name of the perpendiculars:
• For both the triangles, the perpendiculars are named as BD.
   ♦ D is the foot of the perpendicular.
3. Length of the perpendiculars:
• For both the triangles, the length of the perpendiculars are denoted as h.
   ♦ h is called the altitude of the triangle
4. Writing a in terms of b, c and A:
I. In fig.3.40(i), considering right triangle BCD, we have:
BC² = BD² + CD²
⇒ BC² = BD² + (AD - AC)²
⇒ BC² = BD² + AD² + AC² - 2AD.AC
⇒ BC² = AB² + AC² - 2AD.AC
(In the right triangle ABD, BD² + AD² = AB²)
⇒ BC² = AB² + AC² - 2AC.AB cos A
(In the right triangle ABD, AD = AB cos A)
⇒ a² = c² + b² - 2b.c cos A
⇒ a² = b² + c² - 2bc cos A
II. In fig.3.40(ii) also, we can obtain the same result by modifying the steps slightly. Considering right triangle BCD, we have:
BC² = BD² + CD²
⇒ BC² = BD² + (AC - AD)²
⇒ BC² = BD² + AC² + AD² - 2AC.AD
⇒ BC² = BD² + AD² + AC² - 2AC.AD
⇒ BC² = AB² + AC² - 2AD.AC
(In the right triangle ABD, BD² + AD² = AB²)
⇒ BC² = AB² + AC² - 2AC.AB cos A
(In the right triangle ABD, AD = AB cos A)
⇒ a² = c² + b² - 2b.c cos A
⇒ a² = b² + c² - 2bc cos A
◼ We see that:
Relation between a, b, c and cos A is the same for both the triangles.
So the relation is applicable to any triangle. 
5. In the above steps, we did not come across angles B and C. The following steps (6) and (7) will help us to derive a relation involving angle B.
6. Fig.3.41 below shows the same triangles. But this time, the perpendiculars are drawn from vertex C.

Fig.3.41

7. Writing b in terms of c, a and B:
I. In fig.3.40(i), considering right triangle BCD, we have:
AC² = AD² + CD²
⇒ AC² = (AB - BD)² + CD²
⇒ AC² = AB² + BD² - 2AB.BD + CD²
⇒ AC² = AB² + (BD² + CD²) - 2AB.BD
⇒ AC² = AB² + BC² - 2AB.BD
(In the right triangle BCD, BD² + CD² = BC²)
⇒ AC² = AB² + BC² - 2AB.BC cos B
(In the right triangle BCD, BD = BC cos B)
⇒ b² = c² + a² - 2c.a cos B
⇒ b² = c² + a² - 2ca cos B
II. In fig.3.40(ii) also, we can obtain the same result without any modification of the steps. This is because, the arrangements of inner triangles in both cases are similar. The steps will be exactly the same as in 7.I. We will get:
b² = c² + a² - 2ca cos B
◼ We see that:
Relation between b, c, a and cos B is the same for both the triangles.
So the relation is applicable to any triangle.
8. By drawing perpendicular from the vertex A, we will be able to prove the third result:
c² = a² + b² - 2ab cos C
• The reader may write all the steps in his/her own note books.
Hint: The figures will be as shown in fig.3.43 below:

Fig.3.43

---

Let us see some examples:
Example 1:
Fig.3.44(i) below shows a triangle ABC in which two sides and an included angle are given. We are asked to find the remaining one side and the remaining two angles.

Fig.3.44

Solution:
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = -- \
\end{eqnarray}$
2. Using the cosine formula, we have:
c² = a² + b² - 2ab cos C
• Substituting the known values, we get:
c² = 16.84² + 11.17² - 2 × 16.84 × 11.17 × cos 95
⇒ c² = 16.84² + 11.17² - 2 × 16.84 × 11.17 × -0.0871 = 441.143
⇒ c = ± 21.003
• Length cannot be -ve. So we get: c = 21 cm
• So the table becomes:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
3. Once we get this additional information, it is always easier to proceed using sine formula.
$\begin{eqnarray}
&{}& \frac{\sin C}{c}=\frac{\sin B}{b} \\
&\Rightarrow& \frac{\sin 95}{21}=\frac{\sin B}{11.17} \\
&\Rightarrow& \sin B=0.5299 \\
&\Rightarrow& B=31.99=32^{\text{o}} \
\end{eqnarray}$
• Consider the identity: sin x = sin (180-x). We get:
    ♦ sin 32 = sin (180-32) = sin 148 = 0.6952
    ♦ That means, sin A = 0.5299 will give two values for A: 32 and 148
• In our present case, angle at C is already an obtuse angle. No triangle can have two obtuse angles. So the appropriate value for A is 32
• Now the table becomes:
$\begin{eqnarray}
A = -- \,&{|}& B = 32&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
4. Using the sine formula again, we have:
$\frac{\sin A}{a}=\frac{\sin C}{c}$
$\begin{eqnarray}
&{}& \frac{\sin A}{16.84}=\frac{\sin 95}{21} \\
&\Rightarrow& A=53.02=53^{\text{o}} \
\end{eqnarray}$
• Here also we must discard (180-53) because, angle at C is already obtuse.
• Check: A = [180 - (95+32)=53]
• So the final table is:
$\begin{eqnarray}
A = 53 \,&{|}& B = 32&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
5. Note that in this problem, the only option to start calculations, is by using cosine formula. We cannot start with sine formula because, we are given two sides and the included angle.

---

Let us solve this problem using Cosine formula alone.
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = -- \
\end{eqnarray}$
2. Using the Cosine formula, we have:
c² = a² + b² - 2ab cos C
• Substituting the known values, we get:
c² = 16.84² + 11.17² - 2 × 16.84 × 11.17 × cos 95
⇒ c² = 16.84² + 11.17² - 2 × 16.84 × 11.17 × -0.0871 = 441.143
⇒ c = ± 21.003
• Length cannot be -ve. So we get: c = 21 cm
• So the table becomes:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
3. Using the Cosine formula again, we have:
a² = b² + c² - 2bc cos A
• Substituting the known values, we get:
16.84² = 11.17² + 21² - 2 × 11.17 × 21 × cos A
⇒ cos A = 0.6013
⇒ A = 53.03 = 53
• Consider the identity: cos x = cos (360-x). We get:
    ♦ cos 53.03 = cos (360-53) = cos 307 = 0.6013
    ♦ That means, cos C = 0.6013 will give two values for C: 53 and 307
• But 307 is not acceptable because, it is greater than 180
• So the table becomes:
$\begin{eqnarray}
A = 53 \,&{|}& B = --&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
4. Using the Cosine formula again, we have:
b² = a² + c² - 2ac cos B
• Substituting the known values, we get:
11.17² = 16.84² + 21² - 2 × 16.84 × 21 × cos B
⇒ cos B = 0.8481
⇒ B = 31.99 = 32
• Consider the identity: cos x = cos (360-x). We get:
    ♦ cos 32 = cos (360-32) = cos 328 = 0.8481
    ♦ That means, cos C = 0.8481 will give two values for C: 32 and 328
• But 328 is not acceptable because, it is greater than 180
• So the table becomes:
$\begin{eqnarray}
A = 53.03 \,&{|}& B = 32&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$

Example 2:
Fig.3.44(ii) above shows a triangle ABC in which all three sides are given. We are asked to find the three angles.
Solution:
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = -- \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
2. Using the cosine formula, we have:
$\cos A=\frac{b^2+c^2-a^2}{2bc}$
• Substituting the known values, we get:
$\cos A=\frac{3.04^2+5.1^2-3.6^2}{2\times 3.04 \times 5.1}$
⇒ A = 44.04
• So the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = --&{|}& C = -- \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
3. Once we get this additional information, it is always easier to proceed using sine formula.
$\begin{eqnarray}
&{}& \frac{\sin A}{a}=\frac{\sin B}{b} \\
&\Rightarrow& \frac{\sin 44.04}{3.6}=\frac{\sin B}{3.04} \\
&\Rightarrow& \sin B=0.5870 \\
&\Rightarrow& B=35.96^{\text{o}} \
\end{eqnarray}$
• Consider the identity: sin x = sin (180-x). We get:
    ♦ sin 35.96 = sin (180-35.96) = sin 144.04 = 0.5870
    ♦ That means, sin B = 0.5870 will give two values for B: 35.96 and 144.04
• In our present case, we have already calculated A as 44.04
    ♦ If we choose B = 144.04, then (A+B) will become (44.04+144.04) = 188.08
    ♦ The sum cannot be greater than 180.
    ♦ So we must discard 144.04 and choose 35.96
• Now the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.96&{|}& C = -- \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
4. Using the sine formula again, we have:
$\frac{\sin A}{a}=\frac{\sin C}{c}$
$\begin{eqnarray}
&{}& \frac{\sin 44.04}{3.6}=\frac{\sin C}{5.1} \\
&\Rightarrow& \sin C=0.9848 \\
&\Rightarrow& C=80^{\text{o}} \
\end{eqnarray}$
• Consider the identity: sin x = sin (180-x). We get:
    ♦ sin 80 = sin (180-80) = sin 100 = 0.9848
    ♦ That means, sin C = 0.9848 will give two values for C: 80 and 100
• In our present case, we have already calculated A and B as 44.04 and 35.96
    ♦ If we choose C = 80, then (A+B+C) will become (44.04+35.96+80) =160
    ♦ The sum cannot be less than 180.
    ♦ So we must discard 80 and choose 100
• Check: C = [180 - (44.04+35.96)=100]
• So the final table is:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.95&{|}& C = 100 \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
5. Note that in this problem, the only option to start calculations, is by using cosine formula. We cannot start with sine formula because, we are given the three sides.

---

Let us solve this problem using Cosine formula alone.
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = -- \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
2. Using the Cosine formula, we have:
$\cos A=\frac{b^2+c^2-a^2}{2bc}$
• Substituting the known values, we get:
$\cos A=\frac{3.04^2+5.1^2-3.6^2}{2\times 3.04 \times 5.1}$
⇒ A = 44.04
• So the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = --&{|}& C = -- \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
3. Using the Cosine formula again, we have:
b² = a² + c² - 2ac cos B
• Substituting the known values, we get:
3.04² = 3.6² + 5.1² - 2 × 3.6 × 5.1 × cos B
⇒ cos B = 0.8096
⇒ B = 35.94
• Consider the identity: cos x = cos (360-x). We get:
    ♦ cos 35.94 = cos (360-35.94) = cos 324.06 = 0.8096
    ♦ That means, cos B = 0.8096 will give two values for C: 35.94 and 324.06
• But 324.06 is not acceptable because, it is greater than 180
• So the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.94&{|}& C = -- \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
4. Using the Cosine formula again, we have:
c² = a² + b² - 2ab cos C
• Substituting the known values, we get:
5.1² = 3.6² + 3.04² - 2 × 3.6 × 3.04 × cos C
⇒ cos C = -0.1739
⇒ C = 100.02 = 100
• Consider the identity: cos x = cos (360-x). We get:
    ♦ cos 100 = cos (360-100) = cos 260 = -0.1739
    ♦ That means, cos C = -0.1739 will give two values for C: 100 and 260
• But 260 is not acceptable because, it is greater than 180
• So the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.94&{|}& C = 100 \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$

---

In the next section, we will see Napier's Analogies.

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Chapter 3.20 - The Sine Formula

In the previous section, we completed a discussion on trigonometric identities and trigonometric equations. In this section, we will see sine formula.

• Consider the triangle ABC in fig.3.39 below.
• For this triangle, we can define three angles:
   ♦ Angle A is the angle between sides AB and AC 
   ♦ Angle B is the angle between sides BA and BC 
   ♦ Angle C is the angle between sides CA and CB 
• We can define three sides also:
   ♦ a is the length of the side BC (The side opposite angle A)
   ♦ b is the length of the side AC (The side opposite angle B)
   ♦ c is the length of the side AB (The side opposite angle C)

Fig.3.39

◼ Based on the above notations, we can write the sine formula as:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

• For proving this formula, we will analyze an obtuse triangle and an acute triangle.

• An obtuse triangle is that triangle in which one angle is obtuse and the remaining two are acute. (Remember that, no triangle can have more than one obtuse angles). • An acute triangle is that triangle in which all three angles are acute.

• Both triangles can be analyzed using the same steps.
• When the formula is proved for both those types of triangles, we can confidently apply it to any triangle.

The proof can be written in 12 steps:
1. Dropping perpendiculars from the top vertex:
• For the obtuse triangle in fig.3.40(i) below, the perpendicular dropped from the top vertex B can meet the base AC only if we extend the base a bit towards the right.

Fig.3.40

• For the acute triangle in fig.3.40(ii), the perpendicular dropped from the top vertex B can meet the base AC with out any difficulty.
2. Name of the perpendiculars:
• For both the triangles, the perpendiculars are named as BD.
   ♦ D is the foot of the perpendicular.
3. Length of the perpendiculars:
• For both the triangles, the length of the perpendiculars are denoted as h.
   ♦ h is called the altitude of the triangle
4. Writing h in terms of A and c:
I. In fig.3.40(i), considering right triangle ABD, we have:
$\sin A = \frac{h}{c}\; \Rightarrow \; h=c \sin A$
II. In fig.3.40(ii) also, considering right triangle ABD, we have:
$\sin A = \frac{h}{c}\; \Rightarrow \; h=c \sin A$
5. Writing h in terms of a and C:
I. In fig.3.40(i), considering right triangle BCD, we have:
$\sin (180-C) = \frac{h}{a}\; \Rightarrow \; h=a \sin (180-C)$
• Using identity 9.d, we have: sin (180-C) = sin C
   ♦ For example,
   ♦ sin 30 = sin (180-30) = sin 120 = 0.5
   ♦ That means,
         ✰ taking the sine of an angle
         ✰ has the same effect as
         ✰ taking the sine of the supplement of that angle and vice versa.
• Thus we get: $h=a \sin C$
II. In fig.3.40(ii) also, considering right triangle BCD, we have:
$\sin C = \frac{h}{a}\; \Rightarrow \; h=a \sin C$
6. Relation between h, sin A, a, sin C, c:
I. For the obtuse triangle in fig(i):
• Using the results in 4.I and 5.I, we get:
h = c sin A = a sin C
$\Rightarrow \frac{\sin A}{a}=\frac{\sin C}{c}$
II. For the acute triangle in fig(ii) also:
• Using the results in 4.II and 5.II, we get:
h = c sin A = a sin C
$\Rightarrow \frac{\sin A}{a}=\frac{\sin C}{c}$
◼ We see that:
Relation between h, sin A, a, sin C and c is the same for both the triangles.
So the relation is applicable to any triangle. 
7. In the above steps, we did not come across angle B and side b. The following steps from (8) to (11) will help us to bring in those items also.
8. Fig.3.41 below shows the same triangles. But this time, the perpendiculars are drawn from vertex C.

Fig.3.41

9. Writing h in terms of B and a:
I. In fig.3.41(i), considering right triangle BDC, we have:
$\sin B = \frac{h}{a}\; \Rightarrow \; h=a \sin B$
II. In fig.3.41(ii) also, considering right triangle BDC, we have:
$\sin B = \frac{h}{a}\; \Rightarrow \; h=a \sin B$
10. Writing h in terms of b and A:
I. In fig.3.41(i), considering right triangle ACD, we have:
$\sin A = \frac{h}{b}\; \Rightarrow \; h=b \sin A$
II. In fig.3.40(ii) also, considering right triangle ACD, we have:
$\sin A = \frac{h}{b}\; \Rightarrow \; h=b \sin A$
11. Relation between h, sin A, a, sin B, b:
I. For the obtuse triangle in fig.3.41(i):
• Using the results in 9.I and 10.I, we get:
h = a sin B = b sin A
$\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}$
II. For the acute triangle in fig(ii) also:
• Using the results in 9.II and 10.II, we get:
h = a sin B = b sin A
$\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}$
◼ We see that:
Relation between h, sin A, a, sin B and b is the same for both the triangles.
So the relation is applicable to any triangle.
12. Combining the results in (6) and (11), we get:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

---

We see that, the sine formula has three parts. While solving simple problems, we will need only two parts at a time. Let us see some examples:

Example 1:
Fig.3.42(i) below shows triangle ABC in which one angle (C) and it's opposite side are given. One of the two remaining angles is also given.

Fig.3.42

• In total, we have: two angles and one side.
• We are asked to find the remaining one angle and the remaining two sides.
Solution:
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = 32&{|}& C = 95 \\
a = -- \,&{|}& b = -- &{|}& c = 21 \
\end{eqnarray}$
2. Using the sine formula, we have:
$\begin{eqnarray}
&{}& \frac{\sin B}{b}=\frac{\sin C}{c} \\
&\Rightarrow& \frac{\sin 32}{b}=\frac{\sin 95}{21} \\
&\Rightarrow& b=\frac{\sin 32}{1} \times \frac{21}{\sin 95} \\
&\Rightarrow& b=11.17 \; \text{cm} \
\end{eqnarray}$
• Now the table becomes:
$\begin{eqnarray}
A = -- \,&{|}& B = 32&{|}& C = 95 \\
a = -- \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
3. Using the sine formula again, we have:
$\frac{\sin A}{a}=\frac{\sin C}{c}$
• We have: A = [180 - (95+32)=53]
• Now the table becomes:
$\begin{eqnarray}
A = 53 \,&{|}& B = 32&{|}& C = 95 \\
a = -- \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 53}{a}=\frac{\sin 95}{21} \\
&\Rightarrow& a=\frac{\sin 53}{1} \times \frac{21}{\sin 95} \\
&\Rightarrow& a=16.84 \; \text{cm} \
\end{eqnarray}$
4. We can check the result in (3) using the relation:
$\frac{\sin A}{a}=\frac{\sin B}{b}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 53}{a}=\frac{\sin 32}{11.17} \\
&\Rightarrow& a=\frac{\sin 53}{1} \times \frac{11.17}{\sin 32} \\
&\Rightarrow& a=16.84 \; \text{cm} \
\end{eqnarray}$
• This is same as the result in (3)
• So the final table is:
$\begin{eqnarray}
A = 53 \,&{|}& B = 32&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$

Example 2:
Fig.3.42(ii) above shows triangle ABC in which one angle (C) and it's opposite side are given. One of the two remaining sides is also given.
• So in total, we have: two sides and one angle.
• We are asked to find the remaining one side and the remaining two angles.
Solution:
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = 100 \\
a = 3.6 \,&{|}& b = -- &{|}& c = 5.1 \
\end{eqnarray}$
2. Using the sine formula, we have:
$\begin{eqnarray}
&{}& \frac{\sin A}{a}=\frac{\sin C}{c} \\
&\Rightarrow& \frac{\sin A}{3.6}=\frac{\sin 100}{5.1} \\
&\Rightarrow& \sin A=3.6 \times \frac{\sin 100}{5.1}=0.6952 \\
&\Rightarrow& A=44.04^{\text{o}} \
\end{eqnarray}$
• Consider the identity: sin x = sin (180-x). We get:
    ♦ sin 44.04 = sin (180-44.04) = sin 135.96 = 0.6952
    ♦ That means, sin A = 0.6952 will give two values for A: 44.04 and 135.96
• In our present case, angle at C is already an obtuse angle. No triangle can have two obtuse angles. So the appropriate value for A is 44.04
• Now the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = --&{|}& C = 100 \\
a = 3.6 \,&{|}& b = -- &{|}& c = 5.1 \
\end{eqnarray}$
3. Using the sine formula again, we have:
$\frac{\sin A}{a}=\frac{\sin B}{b}$
• We have: B = [180 - (100+44.04)=35.96]
• Now the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.96&{|}& C = 100 \\
a = 3.6 \,&{|}& b = -- &{|}& c = 5.1 \
\end{eqnarray}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 44.04}{3.6}=\frac{\sin 35.96}{b} \\
&\Rightarrow& b=\frac{\sin 35.96}{1} \times \frac{3.6}{\sin 44.04} \\
&\Rightarrow& b=3.04 \; \text{cm} \
\end{eqnarray}$
4. We can check the result in (3) using the relation:
$\frac{\sin B}{b}=\frac{\sin C}{c}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 35.96}{b}=\frac{\sin 100}{5.1} \\
&\Rightarrow& b=\sin 35.96 \times \frac{5.1}{\sin 100} \\
&\Rightarrow& b=3.04 \; \text{cm} \
\end{eqnarray}$
• This is same as the result in (3)
• So the final table is:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.96&{|}& C = 100 \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$

---

• The sine formula can be used when:
(i) We are given any two angles and any one side.
(ii) We are given any two sides and a non- included angle.
• If we are given two sides and the angle included between those two given sides, we will not be able to apply the sine formula.
• Also, if we are given all the tree sides but no angles, we will not be able to apply the sine formula.
• In such cases, we can use the cosine formula. We will see it in the next section.

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Chapter 3.19 - Miscellaneous Examples

In the previous section, we completed a discussion on trigonometric equations. In this section, we will see some miscellaneous examples based on the various topics that we saw so far in this chapter.

The list of trigonometric identities can be seen here. 

Solved example 3.81
If sin x = $\frac{3}{5}$ and cos y = $-\frac{12}{13}$, where x and y both lie in second quadrant, find the value of sin (x+y).
Solution:
1. We have the identity 7: sin (x+y) = sin x cos y + cos x sin y
So in addition to the given sin x and cos y values, we need cos x and sin y values also.
2. We have:
$\begin{eqnarray}
&{}& \cos x  \\
&=& \sqrt{1-\sin^2 x} \;=\;\sqrt{1-\left(\frac{3}{5} \right)^2}  \\
&=& \sqrt{\frac{16}{25}}\;=\;+\frac{4}{5}\;\; \text{or} \;\; -\frac{4}{5}  \
\end{eqnarray}$
• Given that, x is in the second quadrant. So cosine will be -ve.
   ♦ Thus we get: cos x = $-\frac{4}{5}$
3. We have:
$\begin{eqnarray}
&{}& \sin y  \\
&=& \sqrt{1-\cos^2 y} \;=\;\sqrt{1-\left(-\frac{12}{13} \right)^2}  \\
&=& \sqrt{\frac{25}{169}}\;=\;+\frac{5}{13}\;\; \text{or} \;\; -\frac{5}{13}  \
\end{eqnarray}$
• Given that, x is in the second quadrant. So sine will be +ve.
   ♦ Thus we get: sin y = $\frac{5}{13}$
4. Substituting the values in (1), we get:
$\begin{eqnarray}
&{}& \sin (x+y)  \\
&=& \frac{3}{5} \times -\frac{12}{13}+-\frac{4}{5} \times -\frac{5}{13}  \\
&=& -\frac{36}{65}-\frac{20}{65}  \\
&=& -\frac{56}{65}  \
\end{eqnarray}$

Solved example 3.82
Prove that $\cos 2x \cos \frac{x}{2}\;-\;\cos 3x \cos \frac{9x}{2}\;=\; \sin 5x \sin \frac{5x}{2}$
Solution:
1. Consider the first term on the LHS.
• We can applying identity 20.a to this term.
• Identity 20.a is: $\cos a + \cos b = 2\cos \left(\frac{a+b}{2}\right) 2\cos \left(\frac{a-b}{2}\right)$
• In our present case,
$\left(\frac{a+b}{2}\right) \;=\; 2x \; \text{and}\; \left(\frac{a-b}{2}\right) \;=\; \frac{x}{2}$
• Solving the two equations, we get: $a = \frac{5x}{2}\; \text{and}\; b = \frac{3x}{2}$
• So the first term on the LHS becomes: $\frac{1}{2}\left[\cos \frac{5x}{2}+\cos \frac{3x}{2} \right]$ 
2. Consider the second term on the LHS.
• We can applying identity 20.a to this term also.
• In our present case,
$\left(\frac{a+b}{2}\right) \;=\; \frac{9x}{2} \; \text{and}\; \left(\frac{a-b}{2}\right) \;=\; 3x$
• Solving the two equations, we get: $a = \frac{15x}{2}\; \text{and}\; b = \frac{3x}{2}$
• So the second term on the LHS becomes: $\frac{1}{2}\left[\cos \frac{15x}{2}+\cos \frac{3x}{2} \right]$
3. In effect, the LHS becomes:
$\frac{1}{2}\left[\cos \frac{5x}{2}+\cos \frac{3x}{2} \right]-\frac{1}{2}\left[\cos \frac{15x}{2}+\cos \frac{3x}{2} \right]$ 
=$\frac{1}{2}\cos \frac{5x}{2}+\frac{1}{2}\cos \frac{3x}{2}\;-\frac{1}{2}\cos \frac{15x}{2}-\frac{1}{2}\cos \frac{3x}{2}$ 
=$\frac{1}{2}\cos \frac{5x}{2}\;-\frac{1}{2}\cos \frac{15x}{2}$ 
=$\frac{1}{2}\left[\cos \frac{5x}{2}\;-\;\cos \frac{15x}{2}\right]$
4. Now we can apply identity 20.b: $\cos a - cos b =-2\sin \left(\frac{a+b}{2}\right) 2\sin \left(\frac{a-b}{2}\right)$
In our present case,
$a \;=\; \frac{5x}{2} \; \text{and}\; b \;=\; \frac{15x}{2}$
So the LHS becomes:
$\frac{1}{2}\times -2 \times \left[\sin \left(\frac{\frac{5x}{2}+\frac{15x}{2}}{2}\right)\; \sin \left(\frac{\frac{5x}{2}-\frac{15x}{2}}{2}\right)\right]$ 
= $-1 \times \left[\sin 5x\; \sin \left(-\frac{5x}{2}\right)\right]$ 
= $-1 \times \left[-\sin 5x\; \sin \left(\frac{5x}{2}\right)\right]$ 
= $\sin 5x\; \sin \frac{5x}{2}$ = RHS

Solved example 3.83
Find the value of $\tan \frac{\pi}{8}$
Solution:
1. Let x = $\frac{\pi}{8}$
2. We have identity 16: $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$
Applying this identity, we get:
$\tan \left( 2 \times \frac{\pi}{8}\right)=\tan \frac{\pi}{8}=\frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}}$
⇒ $1=\frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}}$
3. Let us put 'u' = $\tan \frac{\pi}{8}$.
• Then the result in (2) becomes: $1=\frac{2 u}{1 - u^2}$
⇒ 1 - u² = 2u
⇒ u² + 2u - 1 = 0
• This is a quadratic equation in u. Solving it, we get:
u = (-1+√2) or (-1-√2)
4. That means: $\tan \frac{\pi}{8}$ = (-1+√2) or (-1-√2)
• tangent of $\frac{\pi}{8}$ cannot be -ve because, $\frac{\pi}{8}$ is in the first quadrant.
• Thus we get: $\tan \frac{\pi}{8}$ = (-1+√2) = √2-1

Solved example 3.84
If $\tan x =\frac{3}{4}, \; \pi < x < \frac{3\pi}{2}$, find the value of $\sin \frac{x}{2},\;\cos \frac{x}{2},\,\text{and}\; \tan \frac{x}{2},$
Solution:
1. Given that: $\tan x =\frac{3}{4}$
• We have the identity: 1 + tan²x = sec²x
• Using this identity, we get:
sec x = $\pm \frac{5}{4}$
⇒ cos x = $\pm \frac{4}{5}$
• Given that: $\pi < x < \frac{3\pi}{2}$.
   ♦ That means, x lies in the third quadrant. In the third quadrant, cosine is -ve.
• So we can write: cos x = $-\frac{4}{5}$
2. We have the identity: 1 - cos²x = sin²x
• Using this identity, we get:
sin²x = $1 - \left(-\frac{4}{5}\right)^2$
⇒ sin x = $\pm \frac{3}{5}$
• Given that: $\pi < x < \frac{3\pi}{2}$.
   ♦ That means, x lies in the third quadrant. In the third quadrant, sine is -ve.
• So we can write: sin x = $-\frac{3}{5}$
3. Let $y=\frac{x}{2}$
• We have identity 14: cos 2y = 2cos²y - 1
• Using this identity, we get:
$\cos 2y = \cos \left(2 \times \frac{x}{2} \right)=\cos x =-\frac{4}{5}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
• Picking the last two items, we get:
$-\frac{4}{5}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
$\begin{eqnarray}
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = 1-\frac{4}{5} \\
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = \frac{1}{5} \\
&\Rightarrow& \cos^2 \left(\frac{x}{2} \right) = \frac{1}{10} \\
&\Rightarrow& \cos \left(\frac{x}{2} \right) = \pm \frac{1}{\sqrt{10}} \
\end{eqnarray}$
• Given that: $\pi < x < \frac{3\pi}{2}$.
• Dividing by 2, we get: $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$.
   ♦ That means, $\frac{x}{2}$ lies in the second quadrant. In the second quadrant, cosine is -ve.
• So we can write: $\cos \frac{x}{2} = -\frac{1}{\sqrt{10}}$
4. Next we will find $\sin \frac{x}{2}$
• We have: $\sin^2 \left(\frac{x}{2} \right)= 1- \cos^2 \left(\frac{x}{2} \right)$
$\begin{eqnarray}
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \left(-\frac{1}{\sqrt{10}} \right)^2 \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \frac{1}{10} \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= \frac{9}{10} \\
&\Rightarrow& \sin \left(\frac{x}{2} \right)= \pm \frac{3}{\sqrt{10}} \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the second quadrant. In the second quadrant, sine is +ve.
• So we can write: $\sin \frac{x}{2} = +\frac{3}{\sqrt{10}}$
5. Next we will find $\tan \frac{x}{2}$
• We have: $\tan \left(\frac{x}{2} \right)= \frac{\sin\left(\frac{x}{2} \right)}{\cos \left(\frac{x}{2} \right)}$
$\begin{eqnarray}
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{+\frac{3}{\sqrt{10}}}{-\frac{1}{\sqrt{10}}} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)=+\frac{3}{\sqrt{10}}\times -\frac{\sqrt{10}}{1} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)=-3 \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the second quadrant. In the second quadrant, tangent is indeed -ve.

Solved example 3.85
If $\tan x =-\frac{4}{3}, \; \text{x is in quadrant II}$, find the value of $\sin \frac{x}{2},\;\cos \frac{x}{2},\,\text{and}\; \tan \frac{x}{2},$
Solution:
1. Given that: $\tan x =-\frac{4}{3}$
• We have the identity: 1 + tan²x = sec²x
• Using this identity, we get:
sec x = $\pm \frac{5}{3}$
⇒ cos x = $\pm \frac{3}{5}$
• Given that: x is in the second quadrant.
   ♦ In the second quadrant, cosine is -ve.
• So we can write: cos x = $-\frac{3}{5}$
2. We have the identity: 1 - cos²x = sin²x
• Using this identity, we get:
sin²x = $1 - \left(-\frac{3}{5}\right)^2$
⇒ sin x = $\pm \frac{4}{5}$
• Given that: x is in the second quadrant.
   ♦ In the second quadrant, sine is +ve.
• So we can write: sin x = $+\frac{4}{5}$
3. Let $y=\frac{x}{2}$
• We have identity 14: cos 2y = 2cos²y - 1
• Using this identity, we get:
$\cos 2y = \cos \left(2 \times \frac{x}{2} \right)=\cos x =-\frac{3}{5}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
• Picking the last two items, we get:
$-\frac{3}{5}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
$\begin{eqnarray}
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = 1-\frac{3}{5} \\
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = \frac{2}{5} \\
&\Rightarrow& \cos^2 \left(\frac{x}{2} \right) = \frac{2}{10} \\
&\Rightarrow& \cos \left(\frac{x}{2} \right) = \pm \frac{\sqrt 2}{\sqrt{10}} \
\end{eqnarray}$
• Given that: x is in the second quadrant.
   ♦ This is same as: $\frac{\pi}{2} < x < \pi $.
• Dividing by 2, we get: $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
   ♦ That means, $\frac{x}{2}$ lies in the first quadrant. In the first quadrant, cosine is +ve.
• So we can write: $\cos \frac{x}{2} = +\frac{\sqrt 2}{\sqrt{10}}$
=$\frac{\sqrt 2}{\sqrt 2 \times \sqrt 5}=\frac{1}{\sqrt 5}=\frac{\sqrt 5}{5}$
4. Next we will find $\sin \frac{x}{2}$
• We have: $\sin^2 \left(\frac{x}{2} \right)= 1- \cos^2 \left(\frac{x}{2} \right)$
$\begin{eqnarray}
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \left(-\frac{\sqrt 5}{5} \right)^2 \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \frac{5}{25} \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= \frac{20}{25}= \frac{4 \times 5}{25} \\
&\Rightarrow& \sin \left(\frac{x}{2} \right)= \pm \frac{2 \sqrt 5}{5} \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the first quadrant. In the first quadrant, sine is +ve.
• So we can write: $\sin \frac{x}{2} = +\frac{2 \sqrt 5}{5}$
5. Next we will find $\tan \frac{x}{2}$
• We have: $\tan \left(\frac{x}{2} \right)= \frac{\sin\left(\frac{x}{2} \right)}{\cos \left(\frac{x}{2} \right)}$
$\begin{eqnarray}
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{+\frac{2 \sqrt 5}{5}}{\frac{\sqrt 5}{5}} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)=+\frac{2 \sqrt 5}{5}\times \frac{5}{\sqrt 5} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)=2 \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the first quadrant. In the first quadrant, tangent is indeed +ve.

Solved example 3.86
If $\cos x =-\frac{1}{3}, \; \text{x is in quadrant III}$, find the value of $\sin \frac{x}{2},\;\cos \frac{x}{2},\,\text{and}\; \tan \frac{x}{2},$
Solution:
1. Given that: $\cos x =-\frac{1}{3}$
• We have the identity: 1 - cos²x = sin²x
• Using this identity, we get:
sin x = $\pm \frac{\sqrt 8}{3}$
• Given that: x is in the third quadrant.
   ♦ In the third quadrant, sine is -ve.
• So we can write: sin x = $-\frac{\sqrt 8}{3}$
2. Let $y=\frac{x}{2}$
• We have identity 14: cos 2y = 2cos²y - 1
• Using this identity, we get:
$\cos 2y = \cos \left(2 \times \frac{x}{2} \right)=\cos x =-\frac{1}{3}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
• Picking the last two items, we get:
$-\frac{1}{3}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
$\begin{eqnarray}
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = 1-\frac{1}{3} \\
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = \frac{2}{3} \\
&\Rightarrow& \cos^2 \left(\frac{x}{2} \right) = \frac{2}{6} \\
&\Rightarrow& \cos \left(\frac{x}{2} \right) = \pm \frac{\sqrt 2}{\sqrt{6}} \
\end{eqnarray}$
• Given that: x is in the third quadrant.
   ♦ This is same as: $ \pi < x < \frac{3\pi}{2} $.
• Dividing by 2, we get: $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$.
   ♦ That means, $\frac{x}{2}$ lies in the second quadrant. In the second quadrant, cosine is -ve.
• So we can write: $\cos \frac{x}{2} = -\frac{\sqrt 2}{\sqrt{6}}$
=$-\frac{\sqrt 2}{\sqrt 2 \times \sqrt 3}=-\frac{1}{\sqrt 3}=-\frac{\sqrt 3}{3}$
3. Next we will find $\sin \frac{x}{2}$
• We have: $\sin^2 \left(\frac{x}{2} \right)= 1- \cos^2 \left(\frac{x}{2} \right)$
$\begin{eqnarray}
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \left(-\frac{\sqrt 3}{3} \right)^2 \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \frac{3}{9} \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= \frac{6}{9} \\
&\Rightarrow& \sin \left(\frac{x}{2} \right)= \pm \frac{\sqrt 6}{3} \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the second quadrant. In the second quadrant, sine is +ve.
• So we can write: $\sin \frac{x}{2} = +\frac{\sqrt 6}{3}$
4. Next we will find $\tan \frac{x}{2}$
• We have: $\tan \left(\frac{x}{2} \right)= \frac{\sin\left(\frac{x}{2} \right)}{\cos \left(\frac{x}{2} \right)}$
$\begin{eqnarray}
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{+\frac{\sqrt 6}{3}}{-\frac{\sqrt 3}{3}} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)=+\frac{\sqrt 6}{3}\times -\frac{3}{\sqrt 3} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)=- \sqrt{2} \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the second quadrant. In the second quadrant, tangent is indeed -ve.

Solved example 3.87
If $\sin x =\frac{1}{4}, \; \text{x is in quadrant II}$, find the value of $\sin \frac{x}{2},\;\cos \frac{x}{2},\,\text{and}\; \tan \frac{x}{2},$
Solution:
1. Given that: $\sin x =\frac{1}{4}$
• We have the identity: 1 - sin²x = cos²x
• Using this identity, we get:
cos x = $\pm \frac{\sqrt 15}{4}$
• Given that: x is in the second quadrant.
   ♦ In the second quadrant, cosine is -ve.
• So we can write: cos x = $-\frac{\sqrt 15}{4}$
2. Let $y=\frac{x}{2}$
• We have identity 14: cos 2y = 2cos²y - 1
• Using this identity, we get:
$\cos 2y = \cos \left(2 \times \frac{x}{2} \right)=\cos x =-\frac{\sqrt 15}{4}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
• Picking the last two items, we get:
$-\frac{\sqrt 15}{4}= 2 \cos^2 \left(\frac{x}{2} \right) - 1$
$\begin{eqnarray}
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = 1-\frac{\sqrt 15}{4} \\
&\Rightarrow& 2 \cos^2 \left(\frac{x}{2} \right) = \frac{4 - \sqrt 15}{4} \\
&\Rightarrow& \cos^2 \left(\frac{x}{2} \right) = \frac{4 - \sqrt 15}{8} \\
&\Rightarrow& \cos \left(\frac{x}{2} \right) = \pm \frac{\sqrt {4 - \sqrt 15}}{\sqrt{8}} \
\end{eqnarray}$
• Given that: x is in the second quadrant.
   ♦ This is same as: $ \frac{\pi}{2} < x < \pi $.
• Dividing by 2, we get: $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
   ♦ That means, $\frac{x}{2}$ lies in the first quadrant. In the first quadrant, cosine is +ve.
• So we can write: $\cos \frac{x}{2} = +\frac{\sqrt {4 - \sqrt 15}}{\sqrt{8}}$
= $\frac{(\sqrt {4 - \sqrt 15})\times \sqrt 2}{(\sqrt{8})\times \sqrt 2}=\frac{\sqrt {8 - 2\sqrt 15}}{(2\sqrt{2})\times \sqrt 2}=\frac{\sqrt {8 - 2\sqrt 15}}{4}$
3. Next we will find $\sin \frac{x}{2}$
• We have: $\sin^2 \left(\frac{x}{2} \right)= 1- \cos^2 \left(\frac{x}{2} \right)$
$\begin{eqnarray}
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \left(\frac{\sqrt {8 - 2\sqrt 15}}{4} \right)^2 \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= 1- \left(\frac{8 - 2\sqrt 15}{16} \right) \\
&\Rightarrow& \sin^2 \left(\frac{x}{2} \right)= \frac{8 + 2\sqrt 15}{16} \\
&\Rightarrow& \sin \left(\frac{x}{2} \right)= \pm \frac{\sqrt {8 + 2\sqrt 15}}{4} \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the first quadrant. In the first quadrant, sine is +ve.
• So we can write: $\sin \frac{x}{2} = +\frac{\sqrt {8 + 2\sqrt 15}}{4}$
4. Next we will find $\tan \frac{x}{2}$
• We have: $\tan \left(\frac{x}{2} \right)= \frac{\sin\left(\frac{x}{2} \right)}{\cos \left(\frac{x}{2} \right)}$
$\begin{eqnarray}
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{\frac{\sqrt {8 + 2\sqrt 15}}{4}}{\frac{\sqrt {8 - 2\sqrt 15}}{4}}  \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{\sqrt {8 + 2\sqrt 15}}{4} \times \frac{4}{\sqrt {8 - 2\sqrt 15}} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{\sqrt {8 + 2\sqrt 15}}{\sqrt {8 - 2\sqrt 15}} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{(\sqrt {8 + 2\sqrt 15})(\sqrt {8 + 2\sqrt 15})}{(\sqrt {8 - 2\sqrt 15})(\sqrt {8 + 2\sqrt 15})} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{8 + 2\sqrt 15}{\sqrt {64-4 \times 15}} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{8 + 2\sqrt 15}{\sqrt {4}} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)= \frac{8 + 2\sqrt 15}{2} \\
&\Rightarrow& \tan \left(\frac{x}{2} \right)= 4 + \sqrt 15 \
\end{eqnarray}$
• We have seen that $\frac{x}{2}$ lies in the first quadrant. In the first quadrant, tangent is indeed +ve.

Solved example 3.88
Prove that $\cos^2 x +\cos^2 \left(x+\frac{\pi}{3}\right) +\cos^2 \left(x-\frac{\pi}{3} \right)=\frac{3}{2}$
Solution:
1. Consider the second term of the LHS.
Assume that, there is no square. Then, applying the identity 3, we will get:
$\cos \left(x+\frac{\pi}{3}\right) $
$\begin{eqnarray}
&=& \cos x \cos \frac{\pi}{3}-\sin x \sin \frac{\pi}{3} \\
&=& \cos x \times \frac{1}{2}-\sin x \times \frac{\sqrt{3}}{2} \\
&{}& \text{squaring this, we get:} \\
&=& \frac{\cos^2 x}{4}- \frac{\sqrt{3}\cos x \sin x}{2}+\frac{3 \sin^2 x}{4} \
\end{eqnarray}$
2. The third term can also be modified like this by applying identity 4. We will get:
$\frac{\cos^2 x}{4}+ \frac{\sqrt{3}\cos x \sin x}{2}+\frac{3 \sin^2 x}{4}$
3. Adding all the three terms of the LHS, we get:
LHS
$\begin{eqnarray}
&=& \cos^2 x + \frac{\cos^2 x}{4}- \frac{\sqrt{3}\cos x \sin x}{2}+\frac{3 \sin^2 x}{4}+\frac{\cos^2 x}{4}+ \frac{\sqrt{3}\cos x \sin x}{2}+\frac{3 \sin^2 x}{4} \\
&=& \cos^2 x + \frac{\cos^2 x}{4}+\frac{3 \sin^2 x}{4}+\frac{\cos^2 x}{4}+\frac{3 \sin^2 x}{4} \\
&=& \cos^2 x + \frac{\cos^2 x}{2}+\frac{3 \sin^2 x}{4}+\frac{3 \sin^2 x}{4} \\
&=& \cos^2 x + \frac{\cos^2 x}{2}+\frac{3 \sin^2 x}{2} \\
&=& \frac{3\cos^2 x}{2}+\frac{3 \sin^2 x}{2} \\
&=& \frac{3}{2}(\sin^2 x+\cos^2 x) \\
&=& \frac{3}{2}\times 1 \\
&=& \frac{3}{2}\;=\;RHS \
\end{eqnarray}$

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Link to some more miscellaneous examples is given below:

Solved examples 3.89 to 3.96

---

In the next section, we will see some miscellaneous examples.

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Chapter 3.18 - More Solved Examples on Trigonometric Equations

In the previous section, we saw some solved examples on trigonometric equations. In this section, we will see a few more solved examples.

The list of trigonometric identities can be seen here.

Solved example 3.69
Find the principal solutions and general solution of the equation: sin 2x - sin 4x + sin 6x = 0
Solution:
1. Given that: sin 2x - sin 4x + sin 6x = 0
• Note that, (2+6)/2 = 4. So we will add sin 2x and sin 6x
2. Using identity 20.c, we get:
$\begin{eqnarray}
&{}& \sin 2x + \sin 6x - \sin 4x = 0 \nonumber \\
&\Rightarrow& 2 \sin \left(\frac{2x+6x}{2} \right) \cos \left(\frac{2x-6x}{2} \right) -\sin 4x = 0 \nonumber \\
&\Rightarrow& 2 \sin 4x \cos (-2x) -\sin 4x = 0 \nonumber \\
&\Rightarrow& \sin 4x [2 \cos (-2x) -1] = 0 \nonumber \\
&\Rightarrow& \sin 4x(2 \cos 2x -1) = 0 \nonumber \
\end{eqnarray}$
• We can write:
sin 4x = 0 or (2cos 2x - 1) = 0
3. We will solve this in two parts A and B.
    ♦ In Part A, we will solve sin 4x = 0
    ♦ In Part B, we will solve (2cos 2x - 1) = 0
Part A: sin 4x = 0
• First we will find the principal solutions.
1. Given that: sin 4x = 0
2. We know that 0 is a principal solution.
• That means, we can put '0' in place of x.
3. We will now find another principal solution. It can be done in 4 steps:
(i) We have seen that, x = 0 is a principal solution.
• When the input x is 0, the left side becomes sin 0
• This gives us: sin 4x = sin 0 = 0
(ii) We want another angle such that, it's sine is also 0
• Such an angle can be calculated using identity 9.d: sin (π-x) = sin x
• We get:
$\sin 0 = \sin (\pi - 0) = \sin \pi$
(iii) Using the results in (i) and (ii), we get:
$\sin 4x = \sin 0 = 0=\sin \pi$
(iv) Picking the first and last items in (iii), we get:
$\sin 4x=\sin \pi$
⇒ $4x=\pi$
⇒ $x=\frac{\pi}{4}$
• Thus we get another value for x
• $0 \leq \frac{\pi}{4} < 2\pi$. So $\frac{\pi}{4}$ (45^o) is a principal solution.

• Now we will write the general solution:
1. Given that: sin 4x = 0
• We have to convert this equation into the form: sin x = sin y
• Just now, we saw that: sin 4x = sin 0
2. This is of the form sin x = sin y
   ♦ In the place of 'x', we have 4x
   ♦ In the place of 'y', we have 0
• Now we can apply theorem 1:
sin x = sin y implies x = n𝞹 + (-1)ⁿy, where n ∈ Z
• We get: $4x=n\pi + (-1)^n \times 0$, where n ∈ Z
⇒ $x=\frac{n\pi}{4}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation sin 4x = 10
• Table 3.10 below shows some of the solutions:

Table 3.10

• The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
    ♦ When n = -3,
    ♦ x = $\frac{-3 \times 180}{4}=-135$

Part B: (2cos 2x - 1) = 0
1. This can be rearranged as: 2cos 2x = 1
⇒ cos 2x = $\frac{1}{2}$
2. We know that $\frac{\pi}{6}$ is a principal solution.
• That means, we can put '$\frac{\pi}{6}$' in place of x.
3. We will now find another principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{6}$ is a principal solution.
• When the input x is $\frac{\pi}{6}$, the left side becomes $\cos \frac{\pi}{3}$
• This gives us: cos 2x = $\cos \frac{\pi}{3}=\frac{1}{2}$
(ii) We want another angle such that, it's cosine is also $\frac{1}{2}$
• Such an angle can be calculated using identity 9.g: cos (2π-x) = cos x
• We get:
$\cos \frac{\pi}{3} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\cos 2x = \cos \frac{\pi}{3} = \frac{1}{2} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(iv) Picking the first and last items in (iii), we get:
$\cos 2x=\cos \frac{5\pi}{3}$
⇒ $2x=\frac{5\pi}{3}$
⇒ $x=\frac{5\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150^o) is a principal solution.

• Now we will write the general solution:
1. Given that: cos 2x = $\frac{1}{2}$
• We have to convert this equation into the form: cos x = cos y
• Just now, we saw that: cos 2x = cos $\frac{\pi}{3}$
2. This is of the form cos x = cos y
   ♦ In the place of 'x', we have 2x
   ♦ In the place of 'y', we have $\frac{\pi}{3}$
• Now we can apply theorem 2:
cos x = cos y implies x = 2n𝞹 ± y, where n ∈ Z
• We get: $2x=2n\pi ± \frac{\pi}{3}$, where n ∈ Z
⇒ $x=n\pi ± \frac{\pi}{6}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation cos 2x = $\frac{1}{2}$
• Tables 3.11 below shows some of the solutions:

Table 3.11

• The principal solutions are shown in red color.
• Let us see a sample calculation for the -ve table above:
    ♦ When n = -3,
    ♦ x = (-3 × 180) - 30
    ♦    = -540 - 30
    ♦    = -570
4. We have in total, three tables in this problem. We can choose any one x value from those tables. If we input that x, in the LHS of the given equation sin 2x - sin 4x + sin 6x = 0. It will reduce to zero.
An example:
• Let us input x = -135. We get:
• LHS = sin (2 × -135) - sin (4 × -135) + sin (6 × -135)
        = sin (-270) - sin (-540) + sin (-810)
        = -sin 270 - -sin 540 + -sin 810
        = - -1 + 0 - 1
        = +1 + 0 - 1
        = 0 = RHS

Solved example 3.70
Find the principal solutions and general solution of the equation: 2cos² x+ 3 sin x = 0
Solution:
1. The given equation can be rearranged as follows:
2(1 - sin²x) + 3 sin x = 0
⇒ 2 - 2sin²x + 3 sin x = 0
⇒ 2sin²x - 3 sin x - 2 = 0
2. Let us put a variable 'u' in place of sin x.
• We will get: 2u² - 3u - 2 = 0
• This is a quadratic equation in u. Solving it, we get:
u = $-\frac{1}{2}$ or u = 2
3. So we can write: sin x = $-\frac{1}{2}$ or sin x = 2
• But sin x cannot be '2' because, the maximum possible value of sin x is '1'
• So we need to consider only sin x = $-\frac{1}{2}$
4. We know that $\sin \frac{\pi}{6}=\frac{1}{2}$
• We have identity 9.f: sin (𝞹+x) = -sin x
• Using this identity, we can write: $\sin \left(\pi + \frac{\pi}{6} \right)=-\sin \frac{\pi}{6}$
⇒ $\sin \frac{7\pi}{6} =-\sin \frac{\pi}{6}$
• But $\sin \frac{\pi}{6}\;\text{is}\;\frac{1}{2}$
• So we get: $\sin \frac{7\pi}{6} =-\frac{1}{2}$
⇒ $\sin \frac{7\pi}{6} =-\frac{1}{2}=\sin x$
⇒ $x= \frac{7\pi}{6}$
• $0 \leq \frac{7\pi}{6} < 2\pi$. So $\frac{7\pi}{6}$ (210^o) is a principal solution.
5. We will now find the other principal solution. It can be done in 6 steps:
(i) We have seen that, x=$\frac{7\pi}{6}$ is a principal solution.
• When the input x is $\frac{7\pi}{6}$, the left side becomes sin $\frac{7\pi}{6}$
• This gives us: sin x = sin $\frac{7\pi}{6} = -\frac{1}{2}$
(ii) We want another angle such that, it's sine is also $-\frac{1}{2}$
• Such an angle can be calculated using identities:
    ♦ 9.d: sin (π-x) = sin x
    ♦ 9.h: sin (2π - x) = - sin x
(iii) Using 9.d, we get: $\sin \frac{7\pi}{6}=\sin \left(\pi - \frac{7\pi}{6} \right) = \sin \frac{-1\pi}{6}$
(iv) From (iii), we get: $\sin \frac{7\pi}{6} = \sin \frac{-1\pi}{6}$
• But $\frac{-1\pi}{6}$ is a -ve angle. We want an angle which lies between 0 and 2π.
• So we apply identity 9.h. We get:
$\sin \left(2\pi - \frac{\pi}{6} \right) = -\sin \frac{\pi}{6}$
⇒ $\sin \frac{11\pi}{6} = -\sin \frac{\pi}{6}$
(v) Using (iii) and (iv), we can write:
$\sin \frac{7\pi}{6}=-\sin \frac{\pi}{6}=\sin \frac{11\pi}{6}$
(vi) Thus we get:
$\sin \frac{7\pi}{6}=\sin \frac{11\pi}{6}=\sin x$
⇒ $x=\frac{11\pi}{6}$
• $0 \leq \frac{11\pi}{6} < 2\pi$. So $\frac{11\pi}{6}$ (330^o) is the other principal solution.

• Now we will write the general solution:
1. Given that: sin x = $-\frac{1}{2}$
• We have to convert this equation into the form: sin x = sin y
• Just now, we saw that: sin x = sin $\frac{7\pi}{6}$
2. This is of the form sin x = sin y
   ♦ In the place of 'x', we have x
   ♦ In the place of 'y', we have $\frac{7\pi}{6}$
• Now we can apply theorem 1:
sin x = sin y implies x = n𝞹 + (-1)ⁿy, where n ∈ Z
• We get: $x=n\pi + (-1)^n \times \frac{7\pi}{6}$, where n ∈ Z
• By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation sin x = $-\frac{1}{2}$
• Table 3.12 below shows some of the solutions:

Table 3.12

 • The principal solutions are shown in red color.
• Let us see a sample calculation for the table above:
    ♦ When n = -3,
    ♦ x = (-3 × 180) + (-1)^-3 × 210
    ♦    = -540 - 210
    ♦    = -750
3. We can choose any one x value from the above table. If we input that x, in the LHS of the given equation 2cos² x+ 3 sin x = 0. It will reduce to zero.
An example:
• Let us input x = -150. We get:
• LHS = 2 cos²(-150) + 3 sin (-150)
        = $2 \times \left(\frac{\sqrt 3}{2} \right)^2 + 3 \times -\frac{1}{2}$
        = $2 \times \frac{3}{4} + 3 \times -\frac{1}{2}$
        = $\frac{3}{2} - \frac{3}{2}$
        = 0 = RHS

Solved example 3.71
Find the principal solutions and general solution of the equation: sin² x - cos x = $\frac{1}{4}$
Solution:
1. The given equation can be rearranged as follows:
1 - cos²x - cos x = $\frac{1}{4}$
⇒ cos²x + cos x - $\frac{3}{4}$ = 0
2. Let us put a variable 'u' in place of cos x.
• We will get: u² + u - $\frac{3}{4}$ = 0
• This is a quadratic equation in u. Solving it, we get:
u = $\frac{1}{2}$ or u = $-\frac{3}{2}$
3. So we can write: cos x = $\frac{1}{2}$ or cos x = $-\frac{3}{2}$
• But cos x cannot be '$-\frac{3}{2}$' because, the least possible value of cos x is '-1'
• So we need to consider only cos x = $\frac{1}{2}$
4. We know that $\cos \frac{\pi}{3}=\frac{1}{2}$
• So we get:
$\cos \frac{\pi}{3} =\frac{1}{2}=\cos x$
⇒ $x= \frac{\pi}{3}$
• $0 \leq \frac{\pi}{3} < 2\pi$. So $\frac{\pi}{3}$ (60^o) is a principal solution.
5. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, x = $\frac{\pi}{3}$ is a principal solution.
If we input x = $\frac{\pi}{3}$, the left side becomes: cos $\frac{\pi}{3}$
• This gives us: cos 2x = $\cos \frac{\pi}{3}=\frac{1}{2}$
(ii) We want another angle such that, it's cosine is also $\frac{1}{2}$
• Such an angle can be calculated using identitiy 9.g: cos (2π-x) = cos x
• We get:
$\cos \frac{\pi}{3} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\cos x = \cos \frac{\pi}{3} = \frac{1}{2} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(iv) Picking the first and last items in (iii), we get:
$\cos x=\cos \frac{5\pi}{3}$
⇒ $x=\frac{5\pi}{3}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{3} < 2\pi$. So $\frac{5\pi}{3}$ (300^o) is a principal solution.

• Now we will write the general solution:
1. Given that: cos x = $\frac{1}{2}$
• We have to convert this equation into the form: cos x = cos y
• Just now, we saw that: cos x = cos $\frac{\pi}{3}$
2. This is of the form cos x = cos y
   ♦ In the place of 'x', we have x
   ♦ In the place of 'y', we have $\frac{\pi}{3}$
• Now we can apply theorem 2:
cos x = cos y implies x = 2n𝞹 ± y, where n ∈ Z
• We get: $x=2n\pi ± \frac{\pi}{3}$, where n ∈ Z
• By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation cos x = $\frac{1}{2}$
• Tables 3.13 below shows some of the solutions:

Table 3.13

• The principal solutions are shown in red color.
• Let us see a sample calculation for the -ve table above:
    ♦ When n = -3,
    ♦ x = (2 × -3 × 180) - 60
    ♦    = -1080 - 60
    ♦    = -1140
3. We can choose any one x value from the above tables. If we input that x, in the LHS of the given equation sin² x - cos x = $\frac{1}{4}$, It will reduce to $\frac{1}{4}$.
An example:
• Let us input x = -300. We get:
• LHS = sin² (-300) - cos (-300)
        = [sin (-300) × sin (-300)] - cos (-300)
        = [-sin 300 × -sin 300] - cos 300
        = [-sin 300 × -sin 300] - cos 300
        = $\left[\frac{\sqrt 3}{2} \times \frac{\sqrt 3}{2} \right]-\frac{1}{2}$
        = $\left[\frac{3}{4}\right]-\frac{1}{2}$
        = $\frac{1}{4}$ = RHS

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Link to some more solved examples is given below:

Solved examples 3.72 to 3.80

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In the next section, we will see some miscellaneous examples.

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Chapter 3.17 - Solved Examples on Trigonomeric Equations

In the previous section, we completed a discussion on the three theorems. In this section, we will see how they can be used to solve trigonometric equations.

• The list of trigonometric identities can be seen here.

Solved example 3.63
Find the solution of the trigonometric equation: sin x = $\frac{1}{2}$
Solution:
1. First we have to convert this equation into the form: sin x = sin y
• So we must find a 'y' such that sin y = $\frac{1}{2}$
• We have: $\sin \frac{\pi}{6} = \frac{1}{2}$
   ♦ So we can write $\sin \frac{\pi}{6}$ in the place of $\frac{1}{2}$
• Thus the given equation becomes:
$\sin x=\sin \frac{\pi}{6}$
2. This is of the form sin x = sin y
   ♦ In the place of 'y', we have $\frac{\pi}{6}$
• Now we can apply theorem 1:
sin x = sin y implies x = nπ + (-1)n y, where n ∈ Z
• We get: $x=n\pi+(-1)^n \frac{\pi}{6}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation sin x = $\frac{1}{2}$
Let us see some random examples:
(i) Let us put n = 5
• We get:
$\begin{eqnarray}
&{}& x=5\pi+(-1)^{5} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=5\pi+(-1)^{5} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=5\pi+(-1) \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=5\pi - \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=\frac{29\pi}{6} \nonumber \
\end{eqnarray}$
• ${\frac{29\pi}{6}}^c \; \text{is}\; 870^o$
The reader may verify that, sin 870 is $\frac{1}{2}$
(ii) Let us put n = -2
• We get:
$\begin{eqnarray}
&{}& x=-2\pi+(-1)^{-2} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=-2\pi+\frac{1}{(-1)^2} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=-2\pi+\frac{1}{1} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=-2\pi+\frac{\pi}{6} \nonumber \\
&\Rightarrow& x=-\frac{11\pi}{6} \nonumber \
\end{eqnarray}$
• ${-\frac{11\pi}{6}}^c \; \text{is}\; -330^o$
The reader may verify that, sin (-330) is $\frac{1}{2}$
4. In this way, infinite number of x values are possible. All 'values of x' thus obtained will satisfy the equation sin x = $\frac{1}{2}$
• Often in scientific and engineering problems, we will want only those 'values of x' which lie in between 0 and 2π.
• For that, we put suitable values of n
5. Let us put suitable values of n in our present case.
We have: $x=n\pi+(-1)^n \frac{\pi}{6}$
(i) Let us put n = 0
• We get:
$\begin{eqnarray}
&{}& x=0 \times \pi+(-1)^{0} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=0+1 \times \frac{\pi}{6} \nonumber \\
&{}& \text{(Any number raised to zero is 1)} \nonumber \\
&\Rightarrow& x=\frac{\pi}{6} \nonumber \\
&{}& 0\leq\frac{\pi}{6}<2\pi \nonumber \
\end{eqnarray}$
• So $\frac{\pi}{6}$ (30^o) is acceptable.
(ii) Let us put n = 1
• We get:
$\begin{eqnarray}
&{}& x=1 \times \pi+(-1)^{1} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=\pi -1 \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=\frac{5\pi}{6} \nonumber \\
&{}& 0\leq\frac{5\pi}{6}<2\pi\nonumber \
\end{eqnarray}$
• So $\frac{5\pi}{6}$ (150^o) is acceptable.
(iii) Let us put n = 2
• We get:
$\begin{eqnarray}
&{}& x=2 \times \pi+(-1)^{2} \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=2\pi +1 \times \frac{\pi}{6} \nonumber \\
&\Rightarrow& x=\frac{13\pi}{6} \nonumber \\
&{}& \frac{13\pi}{6}>2\pi\nonumber \
\end{eqnarray}$
• So $\frac{13\pi}{6}$ (390^o) is not acceptable.
6. Thus the values that lie between 0 and 2π are $\frac{\pi}{6} \; \text{and}\; \frac{5\pi}{6}$
◼ Solutions (values of x) that lie between 0 and 2π are called principal solutions.
• So in our present case, the principal solutions are $\frac{\pi}{6} \; \text{and}\; \frac{5\pi}{6}$
7. The table 3.4 below shows the values of x obtained for various values of n.

Table 3.4

• We see that:
    ♦ When n becomes more and more -ve, x also becomes more and more -ve.
    ♦ When n becomes more and more +ve, x also becomes more and more +ve.
    ♦ The principal solutions are obtained when n = 0 and n= 1
        ✰ They are shown in red color.
8. An easier method to find principal solutions:
• Once we find one of the principal solutions, the other can be calculated using a suitable identity. This can be demonstrated in 4 steps for our present problem:
(i) We have seen that, $x=\frac{\pi}{6}$ is a principal solution.
• When the input x is $\frac{\pi}{6}$, the left side of the given equation becomes: $\sin \frac{\pi}{6}$
• This gives us: $\sin x = \sin \frac{\pi}{6}=\frac{1}{2}$
(ii) We want another angle such that, it's sine is also $\frac{1}{2}$
• Such an angle can be calculated using identity 9.d: sin (𝞹-x) = sin x
• We get:
$\sin \frac{\pi}{6} = \sin \left(\pi - \frac{\pi}{6}\right) = \sin \frac{5\pi}{6}$
(iii) Using the results in (i) and (ii), we get:
$\sin x = \sin \frac{\pi}{6}=\frac{1}{2}=\sin \frac{5\pi}{6}$
(iv) Picking the first and last items in (iii), we get:
$\sin x=\sin \frac{5\pi}{6}$
⇒ $x=\frac{5\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150^o) is the other principal solution.

Solved example 3.64
Find the principal solutions and general solution of the equation: $\sin x=\frac{\sqrt 3}{2}$
Solution:
• First we will find the principal solutions.
1. Given that $\sin x=\frac{\sqrt 3}{2}$
2. We know that $\frac{\pi}{3}$ (same as 60^o) is a principal solution.
• That means, we can put $\frac{\pi}{3}$ in place of x.
3. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{3}$ is a principal solution.
• When the input x is $\frac{\pi}{3}$, the left side becomes $\sin \frac{\pi}{3}$
• This gives us: $\sin x = \sin \frac{\pi}{3}=\frac{\sqrt 3}{2}$
(ii) We want another angle such that, it's sine is also $\frac{\sqrt 3}{2}$
• Such an angle can be calculated using identity 9.d: sin (𝞹-x) = sin x
• We get:
$\sin \frac{\pi}{3} = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \frac{2\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\sin x = \sin \frac{\pi}{3}=\frac{\sqrt 3}{2}=\sin \frac{2\pi}{3}$
(iv) Picking the first and last items in (iii), we get:
$\sin x=\sin \frac{2\pi}{3}$
⇒ $x=\frac{2\pi}{3}$
• Thus we get another value for x
• $0 \leq \frac{2\pi}{3} < 2\pi$. So $\frac{2\pi}{3}$ (120^o) is the other principal solution.

• Now we will write the general solution:
1. Given that: $\sin x=\frac{\sqrt 3}{2}$
• We have to convert this equation into the form: sin x = sin y
• So we must find a 'y' such that sin y = $\frac{\sqrt 3}{2}$
• We have: $\sin \frac{\pi}{3} = \frac{\sqrt 3}{2}$
   ♦ So we can write $\sin \frac{\pi}{3}$ in the place of $\frac{\sqrt 3}{2}$
• Thus the given equation becomes:
$\sin x=\sin \frac{\pi}{3}$
2. This is of the form sin x = sin y
   ♦ In the place of 'y', we have $\frac{\pi}{3}$
• Now we can apply theorem 1:
sin x = sin y implies x = nπ + (-1)n y, where n ∈ Z
• We get: $x=n\pi+(-1)^n \frac{\pi}{3}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation sin x = $\frac{\sqrt 3}{2}$
• Table 3.5 below shows some of the solutions:

Table 3.5

• The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 180) + (-1)^-3 × 60
          = (-3 × 180) + (-1)^-3 × 60
          = -540 + (-1) × 60
          = -540 + - 60
          = -600

Solved example 3.65
Find the principal solutions and general solution of the equation: $\tan x=-\frac{1}{\sqrt 3}$
Solution:
• First we will find the principal solutions.
1. Given that $\tan x=-\frac{1}{\sqrt 3}$
2. We know that $\tan \frac{\pi}{6}=\frac{1}{\sqrt 3}$
• Using identities 9.d and 9.c, we have: tan (π-x) = -tan x 
• So we can write: $\tan \left(\pi -\frac{\pi}{6}  \right)=-\tan \frac{\pi}{6}$
⇒ $\tan \frac{5\pi}{6}=-\tan \frac{\pi}{6}$
3. But $\tan \frac{\pi}{6}=\frac{1}{\sqrt 3}$
• So the result in (2) becomes: $\tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}$
• We can write: $\tan x = \tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}$
• Thus we get: $x=\frac{5\pi}{6}$
• $0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150^o) is a principal solution.
4. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{5\pi}{6}$ is a principal solution.
• When the input x is $\frac{5\pi}{6}$, the left side becomes $\tan \frac{5\pi}{6}$
• This gives us: $\tan x = \tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}$
(ii) We want another angle such that, it's tangent is also $-\frac{1}{\sqrt 3}$
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
$\tan \frac{5\pi}{6} = \tan \left(\pi + \frac{5\pi}{6}\right) = \tan \frac{11\pi}{6}$
(iii) Using the results in (i) and (ii), we get:
$\tan x = \tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}=\tan \frac{11\pi}{6}$
(iv) Picking the first and last items in (iii), we get:
$\tan x=\tan \frac{11\pi}{6}$
⇒ $x=\frac{11\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{11\pi}{6} < 2\pi$. So $\frac{11\pi}{6}$ (330^o) is a principal solution.
5. So the two principal solutions are:
$\frac{5\pi}{6}\;\; \text{and}\;\;\frac{11\pi}{6}$

• Now we will write the general solution:
1. Given that: $\tan x=-\frac{1}{\sqrt 3}$
• We have to convert this equation into the form: tan x = tan y
• So we must find a 'y' such that tan y = $-\frac{1}{\sqrt 3}$
• Just above, we saw that: $\tan \frac{5\pi}{6} =-\frac{1}{\sqrt 3}$
   ♦ So we can write $\tan \frac{5\pi}{6}$ in the place of $-\frac{1}{\sqrt 3}$
• Thus the given equation becomes:
$\tan x=\tan \frac{5\pi}{6}$
2. This is of the form tan x = tan y
   ♦ In the place of 'y', we have $\frac{5\pi}{6}$
• Now we can apply theorem 3:
tan x = tan y implies x = nπ + y, where n ∈ Z
• We get: $x=n\pi+ \frac{5\pi}{6}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation $\tan x=-\frac{1}{\sqrt 3}$
• Table 3.6 below shows some of the solutions:

Table 3.6

 • The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 180) + 150
          = -540 + 150
          = -540 + - 60
          = -390

Solved example 3.66
Find the principal solutions and general solution of the equation: $\cos x=\frac{1}{2}$
Solution:
• First we will find the principal solutions.
1. Given that $\cos x=\frac{1}{2}$
2. We know that $\frac{\pi}{3}$ (same as 60^o) is a principal solution.
• That means, we can put $\frac{\pi}{3}$ in place of x.
3. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{3}$ is a principal solution.
• When the input x is $\frac{\pi}{3}$, the left side becomes $\cos \frac{\pi}{3}$
• This gives us: $\cos x = \cos \frac{\pi}{3}=\frac{1}{2}$
(ii) We want another angle such that, it's cosine is also $\frac{1}{2}$
• Such an angle can be calculated using identity 9.g: cos (2𝞹-x) = cos x
• We get:
$\cos \frac{\pi}{3} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\cos x = \cos \frac{\pi}{3}=\frac{1}{2}=\cos \frac{5\pi}{3}$
(iv) Picking the first and last items in (iii), we get:
$\cos x=\cos \frac{5\pi}{3}$
⇒ $x=\frac{5\pi}{3}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{3} < 2\pi$. So $\frac{5\pi}{3}$ (300^o) is the other principal solution.

• Now we will write the general solution:
1. Given that: $\cos x=\frac{1}{2}$
• We have to convert this equation into the form: cos x = cos y
• So we must find a 'y' such that cos y = $\frac{1}{2}$
• Just now, we saw that: $\cos \frac{\pi}{3} = \frac{1}{2}$
   ♦ So we can write $\cos \frac{\pi}{3}$ in the place of $\frac{1}{2}$
• Thus the given equation becomes:
$\cos x=cos \frac{\pi}{3}$
2. This is of the form cos x = cos y
   ♦ In the place of 'y', we have $\frac{\pi}{3}$
• Now we can apply theorem 2:
cos x = cos y implies x = 2nπ ± y, where n ∈ Z
• We get: $x=2n\pi \pm \frac{\pi}{3}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation cos x = $\frac{1}{2}$
• Tables 3.7 below shows some of the solutions:

Table 3.7

• The principal solutions are shown in red color.
• Let us see a sample calculation for the first table:
   ♦ When n = -3,
   ♦ x = (2 × -3 × 180) + 60
          = -1080 + 60
          = -1020

Solved example 3.67
Find the principal solutions and general solution of the equation: tan 2x = 1
Solution:
• First we will find the principal solutions.
1. Given that: tan 2x = 1
2. We know that $\frac{\pi}{8}$ (same as 22.5^o) is a principal solution.
• That means, we can put $\frac{\pi}{8}$ in place of x.
3. We will now find another principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{8}$ is a principal solution.
• When the input x is $\frac{\pi}{8}$, the left side becomes $\tan \frac{\pi}{4}$
• This gives us: $\tan 2x = \tan \frac{\pi}{4}=1$
(ii) We want another angle such that, it's tangent is also 1
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
$\tan \frac{\pi}{4} = \tan \left(\pi + \frac{\pi}{4}\right) = \tan \frac{5\pi}{4}$
(iii) Using the results in (i) and (ii), we get:
$\tan 2x = \tan \frac{\pi}{4}=1=\tan \frac{5\pi}{4}$
(iv) Picking the first and last items in (iii), we get:
$\tan 2x=\tan \frac{5\pi}{4}$
⇒ $2x=\frac{5\pi}{4}$
⇒ $x=\frac{5\pi}{8}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{8} < 2\pi$. So $\frac{5\pi}{8}$ (112.5^o) is a principal solution.

• Now we will write the general solution:
1. Given that: tan 2x = 1
• We have to convert this equation into the form: tan x = tan y
• Just now, we saw that: tan 2x = tan $\frac{\pi}{4}$
2. This is of the form tan x = tan y
   ♦ In the place of 'x', we have 2x
   ♦ In the place of 'y', we have $\frac{\pi}{4}$
• Now we can apply theorem 3:
tan x = tan y implies x = nπ + y, where n ∈ Z
• We get: $2x=n\pi +\frac{\pi}{4}$, where n ∈ Z
⇒ $x=\frac{n\pi}{2} +\frac{\pi}{8}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation tan 2x = 1
• Table 3.8 below shows some of the solutions:

Table 3.8

• The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 90) + 22.5
          = -270 + 22.5
          = -247.5

Solved example 3.68
Find the principal solutions and general solution of the equation: $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
Solution:
• First we will find the principal solutions.
1. Given that: $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
• We will write the right side in terms of tan. It can be written as follows:
$\begin{eqnarray}
&{}& -\cot\left(x+\frac{\pi}{3}\right) \nonumber \\
&=& \frac{-\cos \left(x+\frac{\pi}{3}\right)}{\sin \left(x+\frac{\pi}{3}\right)} \nonumber \\
&=& \frac{-\;-\;\sin \left(\frac{\pi}{2}+x+\frac{\pi}{3}\right)\;\; \text{(Using identity 9.a)}}{\cos \left( \frac{\pi}{2}+x+\frac{\pi}{3}\right)\;\; \text{(Using identity 9.b)}} \nonumber \\
&=& \frac{\sin \left(x+\frac{5\pi}{6}\right)}{\cos \left(x+\frac{5\pi}{6}\right)} \nonumber \\
&=& \tan \left(x+\frac{5\pi}{6}\right) \nonumber \
\end{eqnarray}$
2. So the given equation becomes: $\tan 2x=\tan \left(x+\frac{5\pi}{6}\right)$
Thus we get: $2x=x+\frac{5\pi}{6}$
⇒ $x=\frac{5\pi}{6}$
$0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150^o) is a principal solution
3. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{5\pi}{6}$ is a principal solution.
• When the input x is $\frac{5\pi}{6}$, the left side becomes $\tan \frac{5\pi}{3}$
• This gives us: $\tan 2x = \tan \frac{5\pi}{3}=\tan \left(x+\frac{5\pi}{6}\right)$
(ii) We want another angle such that, it's tangent remains the same.
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
$\tan \frac{5\pi}{3} = \tan \left(\pi + \frac{5\pi}{3}\right) = \tan \frac{8\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\tan 2x = \tan \frac{5\pi}{3}=\tan \frac{8\pi}{3}=\tan \left(x+\frac{5\pi}{6}\right)$
(iv) Picking the third and last items in (iii), we get:
$\tan \frac{8\pi}{3}=\tan \left(x+\frac{5\pi}{6}\right)$
⇒ $x=\frac{8\pi}{3}-\frac{5\pi}{6}$
⇒ $x=\frac{11\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{11\pi}{6} < 2\pi$. So $\frac{11\pi}{6}$ (330^o) is a principal solution.

• Now we will write the general solution:
1. Given that: $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
• We have to convert this equation into the form: tan x = tan y
• Just now, we saw that: $\tan 2x=\tan \left(x+\frac{5\pi}{6}\right)$
2. This is of the form tan x = tan y
   ♦ In the place of 'x', we have 2x
   ♦ In the place of 'y', we have $\left(x+\frac{5\pi}{6}\right)$
• Now we can apply theorem 3:
tan x = tan y implies x = nπ + y, where n ∈ Z
• We get: $2x=n\pi + x+\frac{5\pi}{6}$, where n ∈ Z
⇒ $x=n\pi +\frac{5\pi}{6}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
• Table 3.9 below shows some of the solutions:

Table 3.9

• The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 180) + 150
          = -540 + 150
          = -390
4. We can choose any x value from the above table. That x value will satisfy the given equation $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
An example:
Let us input x = -570
• Then the LHS becomes: tan (2 × -570) = tan (-1140) = -tan 1140
= -tan (3 × 360 + 60) = -tan 60 = -√3
• RHS becomes: -cot (-570+60) = -cot (510) = -cot (360 + 150) = -cot 150
= -cot (90 + 60) [Using identities 9.a and 9.b] = -tan 60 = -√3
• Thus we get: LHS = RHS

---

In the next section, we will see a few more solved examples.

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