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 15o (ii) cos 15o
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:
$\begin{eqnarray}
LHS &=& \frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} \nonumber \\
&=& \frac{(\sin 5x + \sin x) - 2\sin 3x}{\cos 5x - \cos x} \nonumber \\
&=&
\frac{2 \sin \left( \frac{5x+x}{2}\right) \cos \left(
\frac{5x-x}{2}\right)-2 \sin 3x}{-2 \sin \left( \frac{5x+x}{2}\right)
\sin \left( \frac{5x-x}{2}\right)} \nonumber \\
&=& \frac{2 \sin 3x \cos 2x-2 \sin 3x}{-2 \sin 3x
\sin 2x} \nonumber \\
&=& \frac{2 \sin 3x (\cos 2x -1)}{-2 \sin 3x
\sin 2x} \nonumber \\
&=& \frac{1- \cos 2x}{\sin 2x} \nonumber \\
&=& \frac{2\sin^2 x}{2 \sin x \cos x} \;\; \text{(Identities 14 and 15)} \nonumber \\
&=& \tan x \nonumber \\
&=& RHS \nonumber \
\end{eqnarray}$
Link to some more solved examples is given below:
In
the next
section, we will see trigonometric equations.
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