Chapter 3.13 - Solved Examples on Trigonometric Identities

In the previous section, we completed the derivation of 21 trigonometric identities. We saw a solved example also. In this section, we will see a few more solved examples. The complete list of the 21 identities can be seen here.

Solved example 3.30
Find the value of (i) sin 15^o (ii)  cos 15^o
Solution:
• '15' must be written in terms of those numbers for which we already know the trigonometric ratios.
• '15' can be written as '45 - 30'
• We already know the sine and cosine ratios of 45 and 30
Part (i):
• We write sin 15 as sin (45 - 30)
• Using identity 8, we get:
$\begin{eqnarray} \sin 15 &=&\sin(45-30) \nonumber \\ &=& \sin 45 \cos 30- \cos 45 \sin 30 \nonumber \\ &=& \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}- \frac{1}{\sqrt{2}} \times \frac{1}{2} \nonumber \\ &=& \frac{\sqrt{3}}{2 \sqrt{2}}- \frac{1}{2\sqrt{2}} \nonumber \\ &=& \frac{\sqrt{3}-1}{2\sqrt{2}} \nonumber \ \end{eqnarray}$

Part (ii):
• We write cos 15 as cos (45 - 30)
• Using identity 4, we get:
$\begin{eqnarray} \cos 15 &=&\cos(45-30) \nonumber \\ &=& \cos 45 \cos 30 + \sin 45 \sin 30 \nonumber \\ &=& \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+ \frac{1}{\sqrt{2}} \times \frac{1}{2} \nonumber \\ &=& \frac{\sqrt{3}}{2 \sqrt{2}}+ \frac{1}{2\sqrt{2}} \nonumber \\ &=& \frac{\sqrt{3}+1}{2\sqrt{2}} \nonumber \ \end{eqnarray}$

Solved example 3.31
Find the value of $\tan \frac{13 \pi}{12}$
Solution:
1. We can write $\tan \frac{13 \pi}{12}$ as $\tan (\frac{12 \pi}{12}+\frac{ \pi}{12})=\tan (\pi+\frac{ \pi}{12})$
2. Using identities 9(e) and 9(f), we get:
$\tan (\pi+\frac{ \pi}{12})=\frac{\sin (\pi+\frac{ \pi}{12})}{\cos (\pi+\frac{ \pi}{12})}=\frac{-\sin \frac{ \pi}{12}}{-\cos \frac{ \pi}{12}}=\tan \frac{ \pi}{12}$
• Thus we get: $\tan \frac{13 \pi}{12}=\tan \frac{\pi}{12}$
3. $\tan \frac{\pi}{12}=\frac{\sin \frac{\pi}{12}}{\cos \frac{\pi}{12}}=\frac{\sin 15}{\cos 15}$
We have already seen the values of sin 15 and cos 15 in the previous solved example 3.30. So we can write:
$\begin{eqnarray} \tan \frac{13 \pi}{12} &=& \frac{\sin 15}{\cos 15}=\frac{\sqrt{3}-1}{2\sqrt{2}}\div \frac{\sqrt{3}+1}{2\sqrt{2}}\nonumber \\ &=& \frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{2\sqrt{2}}{\sqrt{3}+1} \nonumber \\ &=& \frac{\sqrt{3}-1}{\sqrt{3}+1} \nonumber \ \end{eqnarray}$
4. Multiplying both numerator and denominator by (√3-1), we get:
$\begin{eqnarray} \tan \frac{13 \pi}{12} &=& \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\nonumber \\ &=& \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^3-1^2)}=\frac{(3- 2\sqrt{3} + 1)}{3-1} \nonumber \\ &=& \frac{(4- 2\sqrt{3}}{2})=\frac{2(2- \sqrt{3}}{2}) \nonumber \\ &=& 2-\sqrt{3} \nonumber \ \end{eqnarray}$

Another method:
1. Same as step (1) above.
2. Same as step (2) above.
3. $\tan \frac{\pi}{12}$ can be written as: $\tan (\frac{\pi}{4}-\frac{\pi}{6})$
Using identity 11, we get:
$\begin{eqnarray} \tan (\frac{\pi}{4}-\frac{\pi}{6}) &=& \frac{\tan \frac{\pi}{4}- \tan \frac{\pi}{6}}{1+\tan \frac{\pi}{4} \tan \frac{\pi}{6}} \nonumber \\ &=& \frac{1- \frac{1}{\sqrt{3}}}{1+1 \times \frac{1}{\sqrt{3}}} \nonumber \\ &=& \frac{\sqrt{3}-1}{\sqrt{3}+1} \nonumber \ \end{eqnarray}$
4. Same as step (4) above.

Solved example 3.32
Prove that $\frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x + \tan y}{\tan x - \tan y}$
Solution:
1. Consider the LHS. Using identities 7 and 8, we get:

$$\frac{\sin (x+y)}{\sin (x-y)}=\frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y}$$
2. Dividing numerator and denominator by cos x cos y, we get:
$$\frac{\sin (x+y)}{\sin (x-y)}=\frac{\frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y}}{\frac{\sin x \cos y}{\cos x \cos y} - \frac{\cos x \sin y}{\cos x \cos y}}$$
$$\Rightarrow \frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x + \tan y}{\tan x - \tan y}$$

Another method:
Consider the RHS. it can be expanded as:
$\begin{eqnarray} \frac{\tan x + \tan y}{\tan x - \tan y} &=& \frac{\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}}{\frac{\sin x}{\cos x} - \frac{\sin y}{\cos y}} \nonumber \\ &=& \frac{\frac{\sin x \cos y + \cos x \sin y}{\cos x \; cos y}}{\frac{\sin x \cos y - \cos x \sin y}{\cos x \; cos y}} \nonumber \\ &=& \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y} \nonumber \\ &=& RHS \nonumber \ \end{eqnarray}$

Solved example 3.33
Show that tan 3x tan 2x tan x = tan 3x - tan 2x - tan x
Solution:
1. We can write tan 3x as: tan (2x + x)
2. We can expand tan (2x+x) using identity 10. We get:
$\begin{eqnarray} \tan (2x+x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x} \nonumber \\ &\Rightarrow&\tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x} \nonumber \\ &\Rightarrow& \tan 3x - \tan x \tan 2x \tan 3x = \tan 2x + \tan x \nonumber \\ &\Rightarrow& \tan 3x - \tan 2x - \tan x =  \tan x \tan 2x \tan 3x \nonumber \\ &\Rightarrow& \tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x \nonumber \ \end{eqnarray}$

Solved example 3.34
Prove that $\cos \left( \frac{\pi}{4}+x\right)+\cos \left( \frac{\pi}{4}-x\right)=\sqrt{2} \cos x$
Solution:
• On the LHS, we have an addition of two cosines. We can convert it into a multiplication of two cosines using identity 20(a). We get:
$\begin{eqnarray} LHS &=& \cos \left( \frac{\pi}{4}+x\right)+\cos \left( \frac{\pi}{4}-x\right) \nonumber \\ &=& 2\cos \left(\frac{\frac{\pi}{4}+x+\frac{\pi}{4}-x}{2}\right) \cos \left(\frac{\frac{\pi}{4}+x-\frac{\pi}{4}+x}{2}\right) \nonumber \\ &=& 2\cos \left(\frac{\frac{\pi}{2}}{2}\right) \cos \left(\frac{2x}{2}\right) \nonumber \\ &=& 2\cos \frac{\pi}{4} \;\cos x \nonumber \\ &=& 2 \times  \frac{1}{\sqrt{2}} \cos x \nonumber \\ &=& \sqrt{2} \cos x \nonumber \\ &=& RHS \nonumber \ \end{eqnarray}$

Another method:
We can apply the identities 3 and 4 on the LHS. we get:
$\begin{eqnarray} LHS &=& \cos \left( \frac{\pi}{4}+x\right)+\cos \left( \frac{\pi}{4}-x\right) \nonumber \\ &=& \cos \frac{\pi}{4}\; \cos x - \sin \frac{\pi}{4} \;\sin x + \cos \frac{\pi}{4} \; \cos x + \sin \frac{\pi}{4} \; \sin x \nonumber \\ &=& \frac{1}{\sqrt{2}} \times \cos x - \frac{1}{\sqrt{2}} \times \sin x + \frac{1}{\sqrt{2}} \times \cos x + \frac{1}{\sqrt{2}} \times \sin x \nonumber \\ &=& 2 \times \frac{1}{\sqrt{2}} \times \cos x \nonumber \\ &=& \sqrt{2} \cos x \nonumber \\ &=& RHS \nonumber \ \end{eqnarray}$

Solved example 3.35
Prove that $\frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x}=\cot x$
Solution:
• On the numerator of LHS, we have an addition of two cosines. We can convert it into a multiplication of two cosines using identity 20(a).
• On the denominator of LHS, we have a subtraction of one sine from another. We can convert it into a multiplication of two sines using identity 20(d).
• We get:
$\begin{eqnarray} LHS &=& \frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} \nonumber \\ &=& \frac{2 \cos \left( \frac{7x+5x}{2}\right) \cos \left( \frac{7x-5x}{2}\right)}{2 \cos \left( \frac{7x+5x}{2}\right) \sin \left( \frac{7x-5x}{2}\right)} \nonumber \\ &=& \frac{\cos \left( \frac{7x-5x}{2}\right)}{\sin \left( \frac{7x-5x}{2}\right)} \nonumber \\ &=& \frac{\cos \left( \frac{2x}{2}\right)}{\sin \left( \frac{2x}{2}\right)} \nonumber \\ &=& \frac{\cos x}{\sin x} \nonumber \\ &=& \cot x \nonumber \\ &=& RHS \nonumber \ \end{eqnarray}$

Solved example 3.36
Prove that $\frac{\sin 5x -3\sin 3x + \sin x}{\cos 5x - \cos x}=\tan x$
Solution:
• On the numerator of LHS, we have addition and subtraction of sines. We can convert it into a multiplication of sines using identity 20(c).
   ♦ Before applying the identity, we must group them suitably.
   ♦ Note that, (5x + x)/2 = 3x
   ♦ So we must group sin 5x and sin x together
• On the denominator of LHS, we have an subtraction of one cosine from another. We can convert it into a multiplication of two cosines using identity 20(b).
• We get:
$\begin{eqnarray} LHS &=& \frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} \nonumber \\ &=& \frac{(\sin 5x + \sin x) - 2\sin 3x}{\cos 5x - \cos x} \nonumber \\ &=& \frac{2 \sin \left( \frac{5x+x}{2}\right) \cos \left( \frac{5x-x}{2}\right)-2 \sin 3x}{-2 \sin \left( \frac{5x+x}{2}\right) \sin \left( \frac{5x-x}{2}\right)} \nonumber \\ &=& \frac{2 \sin 3x \cos 2x-2 \sin 3x}{-2 \sin 3x \sin 2x} \nonumber \\ &=& \frac{2 \sin 3x  (\cos 2x -1)}{-2 \sin 3x \sin 2x} \nonumber \\ &=& \frac{1- \cos 2x}{\sin 2x} \nonumber \\ &=& \frac{2\sin^2 x}{2 \sin x \cos x} \;\; \text{(Identities 14 and 15)} \nonumber \\ &=& \tan x \nonumber \\ &=& RHS \nonumber \ \end{eqnarray}$

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Link to some more solved examples is given below:

Solved examples 3.37 to 3.48

Solved examples 3.49 to 3.62

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In the next section, we will see trigonometric equations.

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