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

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

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