In the previous section, we saw the basics about trigonometric equations. We saw theorem 1 also. In this section, we will see theorems 2 and 3.
Theorem 2
For any real numbers x and y
cos x = cos y implies x = 2nš¹ ± y, where n ∈ Z
•
First we will write an explanation for this theorem. After that, we
will see the proof. The explanation can be written in 3 steps:
1. Given that x and y are any two real numbers.
• Consider the case when:
♦ cosine of x
♦ is equal to
♦ cosine of y
• We often come across such situations in Scientific and Engineering problems.
• For example: cos 30 is equal to cos 1830
♦ [cos 1830 = cos (5 × 360 + 30) = cos 30]
• Another example: cos 30 is equal to cos 2850
♦ [cos 2850 = cos (7 × 360 + 330) = cos 330]
♦ [cos 330 = cos (180 + 150) = -cos 150 (using identity 9.e)]
♦ [-cos 150 = -cos (180 - 30) = -(-cos 30) = cos 30 (using identity 9.c)]
2. When cos x = cos y, there will be a definite relationship between x and y.
• The relation is: x = 2nš¹ ± y, where n ∈ Z
• This relation can be written in words. It can be written in 4 steps:
(i) First, š¹ is to be multiplied by 'two times an integer n'. The product is 2nš¹
(ii) Then y is to be added to or subtracted from 2nš¹
(iii) When steps (i) and (ii) are completed, we will get x
(iv) Each x and y, which satisfy the relation cos x = cos y will have a unique value of 'n'.
3. Let us check this relation:
• In the equation cos 30 = cos 1830, x = 30 and y = 1830
♦ Then the RHS of the relation will be: 2n × 180 ± 1830
♦ If we put n = -5, the RHS will become:
♦ (2 × -5 × 180 + 1830) = (-1800 + 1830) = 30 = LHS
♦ So when x = 30 and y = 1830, the unique value of n is -5
• In the equation cos 30 = cos 2850, x = 30 and y = 2850
♦ Then the RHS of the relation will be: 2n × 180 ± 2850
♦ If we put n = 8, the RHS will become:
♦ (2 × 8 × 180 - 2850) = (2880 - 2850) = 30 = LHS
♦ So when x = 30 and y = 2850, the unique value of n is 8
(At
present, we do not have to worry about how 'n' is calculated. All we
need to know is that, there will be an unique 'n' for each case)
Now we have a basic idea about the theorem. Let us write the proof. It can be written in 7 steps:
1. If cos x = cos y, we can write: cos x - cos y = 0
• To this, we can apply identity 20.b. We get:
$\begin{eqnarray}
&{}&\cos x \;=\; \cos y \nonumber \\
&\Rightarrow & \cos x - \cos y = 0 \nonumber \\
&\Rightarrow & -2\sin \left( \frac{x+y}{2}\right)\,\sin \left( \frac{x-y}{2}\right) = 0 \nonumber \\
&{}&\text{(Applying identity 20.b)} \nonumber \
\end{eqnarray}$
2. If $-2\sin \left(\frac{x+y}{2}\right)\,\sin \left( \frac{x-y}{2}\right) = 0$,
Either $\sin \left( \frac{x+y}{2}\right) = 0$ or $\sin \left( \frac{x-y}{2}\right) = 0$
• That means:
If cos x = cos y,
Either $\sin \left( \frac{x+y}{2}\right) = 0$ or $\sin \left( \frac{x-y}{2}\right) = 0$
3. Let us check this.
(i) We have seen that cos 30 = cos 1830. Here x = 30 and y = 1830
• So (x+y)/ 2 = (30 + 1830)/2 = 1860/2 = 930
♦ Then sin (x+y)/2 = sin 930 = sin (2 × 360+210) = sin 210 = -1
• Also (x-y)/ 2 = (30 - 1830)/2 = -1800/2 = -900
♦ Then sin (x-y)/2 = sin (-900) = -sin 900 = -sin (2 × 360+180)
♦ = -sin 180 = 0
(ii) We have seen that cos 30 = cos 2850. Here x = 30 and y = 2850
• So (x+y)/ 2 = (30 + 2850)/2 = 2880/2 = 1440
♦ Then sin (x+y)/2 = sin 1440 = sin (4 × 360) = 0
• Also (x-y)/ 2 = (30 - 2850)/2 = -2820/2 = -1410
♦ Then sin (x-y)/2 = sin (-1410) = -sin 1410 = -sin (4 × 360-30)
♦ = -sin [2 × 360+(2 × 360-30)] = -sin (2 × 360-30)
♦ = -(-sin 30) = sin 30 = 0.5
(iii) Let us compare the results:
• In step (i), we see that:
When cos 30 = cos 1830, sin (x-y)/2 becomes zero.
• In step (ii), we see that:
When cos 30 = sin 2850, sin (x+y)/2 becomes zero.
(iv) So the result that we obtained in (2) is confirmed.
4. Consider the first result obtained in (2): $\sin \left( \frac{x+y}{2}\right) = 0$
It can be analyzed in 3 steps:
(i) We see that: sine of (x+y)/2 is zero.
• We know the situations when sine becomes zero:
The angle must be a multiple of š¹
• So we can write:
$\text{If}\;\sin \left( \frac{x+y}{2}\right) = 0,\;\;\text{Then}\;
\left( \frac{x+y}{2}\right) = n \pi$, Where n ∈ Z
(ii) If $\frac{x+y}{2} = n \pi$, then (x+y) = 2nš¹
Thus we get: x = 2nš¹ - y
(iii) So we can write:
If $\sin \left( \frac{x+y}{2}\right) = 0$, then x = 2nš¹ - y
5. Consider the second result obtained in (2): $\sin \left( \frac{x-y}{2}\right) = 0$
It can be analyzed in 3 steps:
(i) We see that: sine of (x-y)/2 is zero.
• We know the situations when sine becomes zero:
The angle must be a multiple of š¹
• So we can write:
$\text{If}\;\sin \left( \frac{x-y}{2}\right) = 0,\;\;\text{Then}\;
\left( \frac{x-y}{2}\right) = n \pi$, Where n ∈ Z
(ii) If $\frac{x-y}{2} = n \pi$, then (x-y) = 2nš¹
Thus we get: x = 2nš¹ + y
(iii) So we can write:
If $\sin \left( \frac{x-y}{2}\right) = 0$, then x = 2nš¹ + y
6. Let us compare the results in (4) and (5). The comparison can be done in steps:
(i) In (4), we have: x = 2nš¹ - y
(ii) In (5), we have: x = 2nš¹ + y
(iii) Multiplication of š¹:
♦ In (i) we see that, š¹ can be multiplied by two times any integer.
♦ In (ii) also, we see that, š¹ can be multiplied by two times any integer.
♦ Combining these, we get:
✰ š¹ can be multiplied by two times any integer n
✰ We get: 2nš¹
(iv) Addition or subtraction of 'y'
♦ In (i) we see that, 'y' is to be subtracted.
♦ In (ii) we see that, 'y' is to be added.
♦ Combining these, we get: ±y
(v) So combining the results in (i) and (ii), we get: x = 2nš¹ ± y
7. Let us write a summary. It can be written in 4 steps:
(i) If cos x = cos y,
Then either $\sin \left( \frac{x+y}{2}\right) = 0$ or $\sin \left( \frac{x-y}{2}\right) = 0$
(ii) Then either x = 2nš¹ + y or x = 2nš¹ - y
(iii) Combining the results in (ii), we get: x = 2nš¹ ± y
(iv) Thus Theorem 2 is proved.
Theorem 3
For any real numbers x and y
tan x = tan y implies x = nš¹ + y, where n ∈ Z
•
First we will write an explanation for this theorem. After that, we
will see the proof. The explanation can be written in 3 steps:
1. Given that x and y are any two real numbers.
• Consider the case when:
♦ tangent of x
♦ is equal to
♦ tangent of y
• We often come across such situations in Scientific and Engineering problems.
• For example: tan 30 is equal to tan 1830
♦ [tan 1830 = tan (5 × 360 + 30) = tan 30]
• Another example: tan 30 is equal to tan 2730
♦ [tan 2730 = tan (7 × 360 + 210) = tan 210]
♦ [tan 210 = tan (180 + 30) = tan 30 (using identities 9.e and 9.f)]
2. When tan x = tan y, there will be a definite relationship between x and y.
• The relation is: x = nš¹ + y, where n ∈ Z
• This relation can be written in words. It can be written in 4 steps:
(i) First, š¹ is to be multiplied by an integer n. The product is nš¹
(ii) Then y is to be added to nš¹
(iii) When steps (i) and (ii) are completed, we will get x
(iv) Each x and y, which satisfy the relation tan x = tan y will have a unique value of 'n'.
3. Let us check this relation:
• In the equation tan 30 = tan 1830, x = 30 and y = 1830
♦ Then the RHS of the relation will be: n × 180 + 1830
♦ If we put n = -10, the RHS will become:
♦ (-10 × 180 + 1830) = (-1800 + 1830) = 30 = LHS
♦ So when x = 30 and y = 1830, the unique value of n is -10
• In the equation tan 30 = tan 2730, x = 30 and y = 2730
♦ Then the RHS of the relation will be: n × 180 + 2730
♦ If we put n = -15, the RHS will become:
♦ (-15 × 180 + 2730) = (-2700 + 2730) = 30 = LHS
♦ So when x = 30 and y = 2730, the unique value of n is -15
(At
present, we do not have to worry about how 'n' is calculated. All we
need to know is that, there will be an unique 'n' for each case)
Now we have a basic idea about the theorem. Let us write the proof. It can be written in steps:
1. If tan x = tan y, we can write: tan x - tan y = 0
We can expand this as follows:
$\begin{eqnarray}
&{}&\tan x \;=\; \tan y \nonumber \\
&\Rightarrow & \tan x - \tan y = 0 \nonumber \\
&\Rightarrow & \frac{\sin x}{\cos x} - \frac{\sin y}{\cos y} = 0 \nonumber \\
&\Rightarrow & \frac{\sin x \, \cos y \;\;-\;\;\sin y \, \cos x}{\cos x \, \cos y} = 0 \nonumber \\
&\Rightarrow & \frac{\sin (x-y)}{\cos x \, \cos y} = 0 \nonumber \\
&{}& \text{(Applying identity 8 to the numerator)} \nonumber \
\end{eqnarray}$
2. Division by zero will give a number which does not exist. So (cos x cos y) cannot be zero.
• Only the numerator sin (x-y) can be zero.
• That means, if tan x = tan y, sin(x-y) = 0
3. Let us check this.
(i) We have seen that tan 30 = tan 1830. Here x = 30 and y = 1830
• So (x-y) = (30 - 1830) = -1800
♦ Then sin (x-y) = sin (-1800) = -sin 1800 = -sin (10 × 180) = 0
(ii) We have seen that tan 30 = tan 2730. Here x = 30 and y = 2730
• So (x-y) = (30 - 2730) = -2700
♦ Then sin (x-y) = sin (-2700) = -sin 2700
♦ = -sin (7 × 360 + 180) = -sin 180 = 0
(iii) Let us compare the results:
• In step (i), we see that:
When tan 30 = tan 1830, sin (x-y) becomes zero.
• In step (ii), we see that:
When tan 30 = tan 2730, then also sin (x-y) becomes zero.
(iv) So the result that we obtained in (2) is confirmed.
4. Consider the result obtained in (2): sin (x-y) = 0
It can be analyzed in 3 steps:
(i) We see that: sine of (x-y) is zero.
• We know the situations when sine becomes zero:
The angle must be a multiple of š¹
• So we can write:
If sin (x-y) = 0, Then (x-y) = nš¹, Where n ∈ Z
• Thus we get: x = nš¹ + y
(ii) So we can write:
If sin (x-y) = 0, then x = nš¹ + y
5. Let us write a summary. It can be written in 4 steps:
(i) If tan x = tan y,
Then sin (x-y) = 0
(ii) Then x = nš¹ + y
(iv) Thus Theorem 3 is proved.
We have proved all the three theorems. In
the next
section, we will see how they can be used to solve trigonometric equations.
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