21.14 - Solution of Exponential and Logarithmic Functions

In the previous section, we saw properties of logarithms. In this section, we will see some solved examples which demonstrate the process of solving exponential and logarithmic equations.

Solved example 21.46
Solve the equation: log(6x) − log(4-x) = log 3
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log (6x) - \log (4-x)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log \left(\frac{6x}{4-x} \right)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{6x}{4-x}}    & {~=~}    &{3}    \\
{~\color{magenta}    4    }    &{\implies}    &{6x}    & {~=~}    &{12 – 3x}    \\
{~\color{magenta}    5    }    &{\implies}    &{9x}    & {~=~}    &{12}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{\frac{4}{3}}    \\
\end{array}$                           

Check:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log (6(4/3)) - \log (4- 4/3)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log (8) - \log (8/3)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{8}{8/3}}    & {~=~}    &{3}    \\
{~\color{magenta}    4    }    &{\implies}    &{3}    & {~=~}    &{3}    \\
\end{array}$

Solved example 21.47
Solve the equation: ln(4 −3x) − ln(7x) = ln(11)
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\ln(4-3x) - \ln(7x)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\ln\left[\frac{4-3x}{7x} \right]}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{4-3x}{7x}}    & {~=~}    &{11}    \\
{~\color{magenta}    4    }    &{\implies}    &{4 – 3x}    & {~=~}    &{77x}    \\
{~\color{magenta}    5    }    &{\implies}    &{80x}    & {~=~}    &{4}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{\frac{1}{20}}    \\
\end{array}$                            
 

Check:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\ln(4-3x) - \ln(7x)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\ln(4-3(1/20)) - \ln(7(1/20))}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\ln(77/20) - \ln(7/20)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\ln(77) - \ln(20) - \ln(7) + \ln(20)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    5    }    &{\implies}    &{\ln(77) - \ln(7)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    6    }    &{\implies}    &{\ln(77/7)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    7    }    &{\implies}    &{\ln(11)}    & {~=~}    &{\ln(11)}    \\
\end{array}$

Solved example 21.48
Solve the equation: log₈ (4x + 1) = −1
Solution:

Check:

Solved example 21.49
Solve the equation: 2e^3y+8 − 11e^5−10y = 0
Solution:

Check:

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We have seen the process of solving exponential and logarithmic equations. The reader is advised to try a large number of practice problems in this category. In the next section, we will see derivatives of exponential and logarithmic functions.

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21.13 - Properties of Logarithm

In the previous section, we saw some basic details about logarithmic functions. In this section, we will see some of their properties.

Property I: $\rm{\log_a p \,=\, \frac{\log_b p}{\log_b a}}$
   ♦ On the L.H.S, we have log of p to the base a.
   ♦ On the R.H.S, we have log of p to the base b.
• So this property gives us the method to connect two logs of the same number when the bases are different.
• This property is also known as change of base rule.
• We can write the proof in 5 steps:

1. Let $\rm{\log_a p = \alpha}$
Then $\rm{a^{\alpha} = p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b a = \gamma}$
Then $\rm{b^{\gamma} = a}$
4. Substituting (3) in (1), we get:
$\rm{\left(b^{\gamma} \right)^{\alpha} = p}$
⇒ $\rm{b^{\gamma \alpha} = p}$
5. Substituting (4) in (2), we get:
$\rm{b^{\beta} = p = b^{\gamma \alpha}}$
⇒ $\rm{\beta = \gamma \alpha}$
⇒ $\rm{\alpha = \frac{\beta}{\gamma}}$
⇒ $\rm{\log_a p = \frac{\log_b p}{\log_b a}}$

Property II: $\rm{\log_b pq \,=\, \log_b p + \log_b q}$
   ♦ On the L.H.S, we have log of a product pq.
   ♦ On the R.H.S, we have addition of two individual logs.
• So this property helps us to simplify logarithmic equations when products are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b pq = \alpha}$
Then $\rm{b^{\alpha} = pq}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\gamma}$
⇒ $\rm{b^\alpha = b^{\beta + \gamma}}$
⇒ $\rm{\alpha = \beta + \gamma}$
⇒ $\rm{\log_b pq = \log_b p + \log_b q}$

Property III: $\rm{\log_b \left(\frac{p}{q} \right) \,=\, \log_b p - \log_b q}$
   ♦ On the L.H.S, we have log of a ratio $\frac{p}{q}$.
   ♦ On the R.H.S, we have subtraction of one individual log from another.
• So this property helps us to simplify logarithmic equations when ratios are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b \left(\frac{p}{q} \right) = \alpha}$
Then $\rm{b^{\alpha} = \frac{p}{q}}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = \frac{b^\beta}{b^\gamma}}$
⇒ $\rm{b^\alpha = b^{\beta - \gamma}}$
⇒ $\rm{\alpha = \beta - \gamma}$
⇒ $\rm{\log_b pq = \log_b p - \log_b q}$

Property IV: $\rm{\log_b p^2 \,=\, 2 \log_b p}$
   ♦ On the L.H.S, we have log of a square.
   ♦ On the R.H.S, we have individual log multiplied by 2.
• So this property helps us to simplify logarithmic equations when squares are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta}$
⇒ $\rm{\log_b (p \times p) = \log_b p + \log_b p}$
⇒ $\rm{\log_b p^2 = 2 \log_b p}$

Property V: $\rm{\log_b p^3 \,=\, 3 \log_b p}$
   ♦ On the L.H.S, we have log of a cube.
   ♦ On the R.H.S, we have individual log multiplied by 3.
• So this property helps us to simplify logarithmic equations when cubes are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta + \beta}$
⇒ $\rm{\log_b (p \times p \times p) = \log_b p + \log_b p + \log_b p}$
⇒ $\rm{\log_b p^3 = 3 \log_b p}$

Property VI: $\rm{\log_b p^n \,=\, n \log_b p}$
   ♦ On the L.H.S, we have log of a n^th power.
   ♦ On the R.H.S, we have individual log multiplied by n.
• So this property helps us to simplify logarithmic equations when n^th power is involved.
• The proof can be written using the principles of mathematical induction.

Property VII: $\rm{\log_p p \,=\, 1}$
   ♦ On the L.H.S, we have same number and base.
   ♦ On the R.H.S, we have 1.
We can write the proof as follows:
• Let $\rm{\log_p p = \alpha}$
• Then $\rm{p^{\alpha} = p}$
⇒ $\rm{p^{\alpha} = p^1}$
⇒ $\rm{\alpha = 1}$

Property VIII: $\rm{\log_b p \,=\, \frac{1}{\log_p b}}$
   ♦ On the L.H.S, we have base b and number p.
   ♦ On the R.H.S, we have base p and number b.
   ♦ So base and number are interchanged.
We can write the proof as follows:
• Applying the change of base rule we get:
$\rm{\log_b p \,=\, \frac{\log_p p}{\log_p b} \,=\, \frac{1}{\log_p b}}$ 

Property IX: $\rm{\log_p {p^x} \,=\, x}$
We can write the proof as follows:
• Let $\rm{\log_p p^x = \alpha}$
• Then $\rm{p^{\alpha} = p^x}$
⇒ $\rm{\alpha = x}$

• Note that, this property is used when we first begin to learn about logarithms. For example:
$\rm{\log_5 625 = \log_5 5^4 = 4}$

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Now we will see some solved examples

Solved example 21.43
Express as a single logarithm:
$\rm{2 \log x - 5 \log y + 3 \log z}$
Solution:
1. 2 log x = log x².
2. −5 log y = −log y⁵.
3. 3 log z = log z³.
4. log x² − log y⁵ + log z³ = $\rm{\frac{x^2 z^3}{y^5}}$

Solved example 21.44
Expand using the properties of logarithm:
$\rm{\log_5 \sqrt[3]{x}}$
Solution:
$\rm{\log_5 \sqrt[3]{x}}$
= $\rm{\log_5 \left(x^{\frac{1}{3}} \right)}$ 
= $\rm{\frac{1}{3} \log_5 x}$

Solved example 21.45
Evaluate log₅ 28
Solution:
1. We could find the solution easily if it is "25" instead of "28".
2. We cannot directly use the calculator because base is 5.
3. So we will apply the change of base rule. We get:
$\rm{\log_5 28 \,=\,\frac{\log_{10} 28}{\log_{10} 5}\,=\,\frac{1.44715}{0.69897}\,=\,2.07040}$

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In the next section, we will see exponential and logarithmic equations.

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Chapter 3.11 - Trigonometric Identities

In the previous section, we saw the domain and range of tangent and cotangent functions. In this section, we will derive some trigonometric identities.

First we will derive two simple identities. They can be derived in 21 steps:
1. In fig.3.34(a) below, a ray has rotated by x^c in the +ve direction.
• Another ray has rotated by the same x^c in the negative direction.

Fig.3.34

2. Since the angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
3. So the coordinates of Q are: (a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = -b and cos (x) = a
4. For the +ve angle x^c, the coordinates of P are: (a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = b and cos x = a
5. comparing the results in (3) and (4), we get:
sin (-x) = -sin x and cos (-x) = cos x

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6. In fig.3.34(b) above, a ray has rotated by x^c in the +ve direction.
    ♦ x is greater than $\frac{\pi}{2}$
• Another ray has rotated by the same x^c in the negative direction.
7. From the fig., it is clear that:
    ♦ ∠MOP = (π - x)
    ♦ ∠MOQ = (π - x)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
8. So the coordinates of Q are: (-a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = -b and cos (-x) = -a
9. For the +ve angle x^c, the coordinates of P are: (-a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = b and cos x = -a
10. comparing the results in (8) and (9), we get:
sin (-x) = -sin x and cos (-x) = cos x

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11. In fig.3.35(a) below, a ray has rotated by x^c in the +ve direction.
    ♦ x is greater than π
• Another ray has rotated by the same x^c in the negative direction.

Fig.3.35

12. From the fig., it is clear that:
    ♦ ∠MOP = (x - π)
    ♦ ∠MOQ = (x - π)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
13. So the coordinates of Q are: (-a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = b and cos (-x) = -a
14. For the +ve angle x^c, the coordinates of P are: (-a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = -b and cos x = -a
15. comparing the results in (13) and (14), we get:
sin (-x) = -sin x and cos (-x) = cos x

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16. In fig.3.35(b) above, a ray has rotated by x^c in the +ve direction.
    ♦ x is greater than $\frac{3\pi}{2}$
• Another ray has rotated by the same x^c in the negative direction.
17. From the fig., it is clear that:
    ♦ ∠MOP = (2π - x)
    ♦ ∠MOQ = (2π - x)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
18. So the coordinates of Q are: (a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = b and cos (-x) = a
19. For the +ve angle x^c, the coordinates of P are: (a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = -b and cos x = a
20. comparing the results in (18) and (19), we get:
sin (-x) = -sin x and cos (-x) = cos x

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21. We have considered the position of the -ve angle in all the four quadrants.
• We get the same result in all the four steps (5), (10), (15) and (20)
• So we can write the first and second identities:
Identity 1: sin (-x) = -sin x
Identity 2: cos (-x) = cos x

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Next we will derive the third identity. It is related to the sum of two angles. It can be derived in 7 steps:
1. In fig.3.36 below, the green circle is a unit circle.

Fig.3.36

• Beginning from the +ve side of the x axis, the red ray rotated by x^c in the anticlockwise direction.
• Beginning from the +ve side of the x axis, the magenta ray rotated by (x+y)^c in the anticlockwise direction.
    ♦ So the angle between the red and magenta rays is y^c
• Beginning from the +ve side of the x axis, the blue ray rotated by y^c in the clockwise direction.
2. Important points and their coordinates:
• P₁ is the tip of the red ray.
    ♦ It's coordinates will be: (cos x, sin x)
• P₂ is the tip of the magenta ray.
    ♦ It's coordinates will be: (cos (x+y), sin (x+y))
• P₃ is the tip of the blue ray.
    ♦ It's coordinates will be: (cos (-y), sin (-y))
• P₄ is the point where the unit circle cuts the +ve side of the x axis.
    ♦ It's coordinates will be: (1, 0)
3. Using the five points O, P₁, P₂, P₃ and P₄, we can form two triangles:
P₁OP₃ and P₂OP₄
These two triangles are congruent. It can be proved in 5 steps:
(i) ∠ P₂OP₃ = (2π - x - y - y) = (2π - x - 2y)
(ii) In the triangle P₁OP₃,
∠ P₁OP₃ = [y + ∠ P₂OP₃] = [y + (2π - x - 2y)] = [2π - x - y]
(iii) In the triangle P₂OP₄,
∠ P₂OP₄ = [∠ P₂OP₃ + y] = [(2π - x - 2y) + y] = [2π - x - y]
(iv) So the same angle [2π - x - y] is present in both the triangles.
(v) In triangle P₁OP₃, this [2π - x - y] angle is included in between two radii.
• In triangle P₂OP₄ also, this [2π - x - y] angle is included in between two radii.
• So it is a SAS congruence.
4. Since the two triangles are congruent, we get: P₁P₃ = P₂P₄
5. Lengths of the these two sides can be calculated using distance formula.

• The distance between two points A (x1, y1) and B (x2, y2) is given by: AB2 = [(x2 - x1)2 + (y2 - y1)2] • Details can be seen here.

◼ Let us first calculate P₁P₃. It can be written in 3 steps:
(i) First we write the coordinates:
    ♦ Coordinates of P₁ are: (cos x, sin x)
    ♦ Coordinates of P₃ are: (cos (-y), sin (-y))
(ii) Now we apply the distance formula:
(P₁P₃)² = [(cos (-y) - cos x)² + (sin (-y) - sin x)²]
(iii) Using identities 1 and 2, we have: sin (-y) = -sin y and cos (-y) = cos y
• So the result in (ii) becomes:
(P₁P₃)² = [(cos y - cos x)² + (-sin y - sin x)²]
⇒ (P₁P₃)² = cos²y - 2 cos x cos y + cos²x + (-sin y)² - 2 (-sin y) (sin x) + sin²x
⇒ (P₁P₃)² = cos²y - 2 cos x cos y + cos²x + sin²y + 2 sin y sin x + sin²x
⇒ (P₁P₃)² = 2 - 2 cos x cos y + 2 sin y sin x
[∵ cos²y + sin²y  = 1 and cos²x + sin²x = 1]
◼ Next let us calculate P₂P₄. It can be written in 2 steps:
(i) First we write the coordinates:
    ♦ Coordinates of P₂ are: (cos (x+y), sin (x+y))
    ♦ Coordinates of P₄ are: (1, 0)
(ii) Now we apply the distance formula:
(P₂P₄)² = [(1 - cos (x+y))² + (0 - sin (x+y))²]
⇒ (P₂P₄)² = [1 - 2 cos (x+y) + cos²(x+y) + 0² - 2 0 sin (x+y) + sin²(x+y)]
⇒ (P₂P₄)² = [1 - 2 cos (x+y) + cos²(x+y) + sin²(x+y)]
⇒ (P₂P₄)² = [2 - 2 cos (x+y)]
[∵ cos²(x+y) + sin²(x+y) = 1]
6. From (4), the two lengths P₁P₃ and P₂P₄ are equal. So their squares must also be equal.
• Equating the squares, we get:
2 - 2 cos x cos y + 2 sin y sin x = 2 - 2 cos (x+y)
⇒ -2 cos x cos y + 2 sin y sin x = -2 cos (x+y)
⇒ -cos x cos y + sin y sin x = -cos (x+y)
⇒ cos x cos y - sin y sin x = cos (x+y)
7. Thus we get the third identity:
Identity 3: cos (x+y) = cos x cos y - sin x sin y

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Next we will derive the fourth identity. It is related to the difference of two angles. It can be derived in 3 steps:
1. We have the Identity 3: cos (x+y) = cos x cos y - sin x sin y
2. Put (-y) in place of y. We get:
cos (x+(-y)) = cos x cos (-y) - sin x sin (-y)   
⇒ cos (x-y) = cos x cos y - sin x × -sin y
[∵ From first and second identities, cos (-y) = cos y and sin (-y) = -sin y]   
3. Thus we get the fourth identity:
Identity 4: cos (x-y) = cos x cos y + sin x sin y

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• Identities 3 and 4 follow a similar pattern. The only difference is in the sign.
• So we can write them together as:
Identities 3 & 4: cos (x±y) = cos x cos y ∓ sin x sin y.
• On the left side, we have ± and on the right side, we have ∓.
   ♦ If on the left side we apply the top sign '+', on the right side also, we must apply the top sign '-'
   ♦ If on the left side we apply the bottom sign '-', on the right side also, we must apply the bottom sign '+'.

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Next we will derive the fifth identity. It can be derived in 3 steps:
1. We have the Identity 4: cos (x-y) = cos x cos y + sin x sin y
2. put x = $\frac{\pi}{2}$ and y = x. We get:
$\cos (\frac{\pi}{2} - x)=\cos \frac{\pi}{2} \sin x + \sin \frac{\pi}{2} \sin x$
$\cos (\frac{\pi}{2} - x)=0 \times \sin x + 1 \times \sin x$
3. Thus we get the fifth identity:
Identity 5: $\cos (\frac{\pi}{2} - x)= \sin x$

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Next we will derive the sixth identity. It can be derived in 3 steps:
1. Using the identity 5, we can write:
$\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{2} - x\right)\right]= \sin \left(\frac{\pi}{2} - x\right)$
2. But $\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{2} - x\right)\right]$ is cos x
So the result in (1) becomes: $\cos x = \sin \left(\frac{\pi}{2} - x\right)$
3. Thus we get the sixth identity:
Identity 6: $ \sin \left(\frac{\pi}{2} - x\right)=\cos x$

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Next we will derive the seventh identity. It can be derived in 4 steps:
1. Using the identity 5, we can write:
$\cos \left[\frac{\pi}{2}-(x+y)\right]= \sin (x+y)$
2. The left side can be rearranged. We get:
$\cos \left[\left(\frac{\pi}{2}-x\right)-y \right]= \sin (x+y)$
3. Using identity 4, we can expand the left side. We get:
$\cos \left(\frac{\pi}{2}-x\right) \, \cos y + \sin \left(\frac{\pi}{2}-x\right) \, \sin y = \sin (x+y)$
⇒ $\sin x \, \cos y + \cos x \sin y = \sin (x+y)$
4. Thus we get the seventh identity:
Identity 7: $\sin (x+y)=\sin x \, \cos y + \cos x \sin y$

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Next we will derive the eighth identity. It can be derived in 2 steps:
1. In identity 7, put -y in place of y. We get:
$\sin (x+(-y))=\sin x \, \cos (-y) + \cos x \sin (-y) $  
⇒ $\sin (x-y)=\sin x \, \cos y + \cos x \times -\sin y $  
⇒ $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
2. Thus we get the eithth identity:
Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $

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• Identities 7 and 8 follow a similar pattern. The only difference is in the sign.
• So we can write them together as:
Identities 7 & 8: sin (x±y) = sin x cos y ± cos x sin y.
• On the left side, we have ± and on the right side, we have ±.
   ♦ If on the left side we apply the top sign '+', on the right side also, we must apply the top sign '+'
   ♦ If on the left side we apply the bottom sign '-', on the right side also, we must apply the bottom sign '-'.

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Next we will derive the ninth identity. Here we will derive 8 simple identities from 9(a) to 9(h). They can be derived in steps:
1. Consider the Identity 3: cos (x+y) = cos x cos y - sin x sin y.
• We put $\frac{\pi}{2}$ in place of x and x in place of y. We get:
$\cos (\frac{\pi}{2}+x) = \cos \frac{\pi}{2}\, \cos x - \sin \, \frac{\pi}{2} \sin x$
⇒ $\cos (\frac{\pi}{2}+x) =0 \times \cos x - 1 \times \sin x$
⇒ $\cos (\frac{\pi}{2}+x) =- \sin x$
Identity 9(a): $\cos (\frac{\pi}{2}+x) =- \sin x$

2. Consider the Identity 7: $\sin (x+y)=\sin x \, \cos y + \cos x \sin y$
• We put $\frac{\pi}{2}$ in place of x and x in place of y. We get:
$\sin (\frac{\pi}{2}+x) = \sin \frac{\pi}{2}\, \cos x + \cos \, \frac{\pi}{2} \sin x$
⇒ $\sin (\frac{\pi}{2}+x) =1 \times \cos x + 0 \times \sin x$
⇒ $\sin (\frac{\pi}{2}+x) = \cos x$
Identity 9(b): $\sin (\frac{\pi}{2}+x) = \cos x$

3. Consider the Identity 4: cos (x-y) = cos x cos y + sin x sin y.
• We put π in place of x and x in place of y. We get:
cos (π - x) = cos π cos x + sin π sin x.
⇒ cos (π - x) = -1 × cos x + 0 × sin x.
⇒ cos (π - x) = - cos x
Identity 9(c): cos (π - x) = - cos x

4. Consider the Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
• We put π in place of x and x in place of y. We get:
sin (π - x) = sin π cos x - cos π sin x
⇒ sin (π - x) = 0 × cos x - (-1)  × sin x
⇒ sin (π - x) = sin x
Identity 9(d): sin (π - x) = sin x

5. Consider the Identity 3: cos (x+y) = cos x cos y - sin x sin y.
• We put π in place of x and x in place of y. We get:
cos (π + x) = cos π cos x - sin π sin x
⇒ cos (π + x) = -1 × cos x - 0 × sin x
⇒ cos (π + x) = -cos x
Identity 9(e): cos (π + x) = -cos x 

6. Consider the Identity 3: sin (x+y) = sin x cos y + cos x sin y.
• We put π in place of x and x in place of y. We get:
sin (π + x) = sin π cos x + cos π sin x.
⇒ sin (π + x) = 0 × cos x + (-1) × sin x
⇒ sin (π + x) = -sin x
Identity 9(f): sin (π + x) = -sin x

7. Consider the Identity 4: cos (x-y) = cos x cos y + sin x sin y.
• We put 2π in place of x and x in place of y. We get:
cos (2π - x) = cos 2π cos x + sin 2π sin x.
⇒ cos (2π - x) = +1 × cos x + 0 × sin x.
⇒ cos (2π - x) = cos x
Identity 9(g): cos (2π - x) = cos x.

8. Consider the Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
• We put 2π in place of x and x in place of y. We get:
sin (2π - x) = sin 2π cos x - cos 2π sin x
⇒ sin (2π - x) = 0 × cos x - 1  × sin x
⇒ sin (2π - x) = -sin x
Identity 9(h): sin (2π - x) = sin x

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• The above identities from 9(a) to 9(h) are related to sin x and cos x. Using them, we can derive similar identities for tan x, cot x, sec x and csc x.
• Let us see an example:
$\tan (\frac{\pi}{2}+x)=\frac{\sin (\frac{\pi}{2}+x)}{\cos (\frac{\pi}{2}+x)}=\frac{\cos x}{\sin x}=\tan x$
• Thus we get the identity: $\tan (\frac{\pi}{2}+x)=\tan x$

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

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