Question

Match the following trigonometric ratios with their corresponding values.

Answer 1

Given : a) $\cos \frac{\pi }{8}$
We know that, $\frac{\pi }{8}=\frac{\pi }{2}-\frac{3\pi }{8}$
$\Rightarrow \cos \frac{\pi }{8}=\cos (\frac{\pi }{2}-\frac{3\pi }{8})$
$\Rightarrow \cos \frac{\pi }{8}=\sin \left(\frac{3\pi }{8}\right)$

Also we know that,
$|\cos \theta |=\sqrt{\frac{1+\cos 2\theta }{2}}....\left(i\right)$
Now, put $\theta =\frac{\pi }{8}$

$\therefore \cos \frac{\pi }{8}=\sqrt{\frac{1+\cos \frac{2\pi }{8}}{2}}$

$\Rightarrow \cos \frac{\pi }{8}=\sqrt{\frac{1+\cos \frac{\pi }{4}}{2}}$

$\Rightarrow \cos \frac{\pi }{8}=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}}$

$\Rightarrow \cos \frac{\pi }{8}=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$

Ratioonalising the R.H.S. we get,
$\Rightarrow \cos \frac{\pi }{8}=\sqrt{\frac{(\sqrt{2}+1)2\sqrt{2}}{2\sqrt{2}\left(2\sqrt{2}\right)}}$

$\Rightarrow \cos \frac{\pi }{8}=\sqrt{\frac{4+2\sqrt{2}}{8}}$
$\Rightarrow \cos \frac{\pi }{8}=\sqrt{\frac{2+\sqrt{2}}{4}}$

$\Rightarrow \cos \frac{\pi }{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$

b) $\tan \frac{\pi }{8}$
We know that, $\frac{\pi }{8}=\frac{\pi }{2}-\frac{3\pi }{8}$
$\Rightarrow \tan \frac{\pi }{8}=\tan (\frac{\pi }{2}-\frac{3\pi }{8})$
$\Rightarrow \tan \frac{\pi }{8}=\cot \left(\frac{3\pi }{8}\right)$

Also we know that,
$|\tan \theta |=\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}$
Now, put $\theta =\frac{\pi }{8}$

$\therefore \tan \frac{\pi }{8}=\sqrt{\frac{1-\cos 2\frac{\pi }{8}}{1+\cos 2\frac{\pi }{8}}}$

$\Rightarrow \tan \frac{\pi }{8}=\sqrt{\frac{1-\cos \frac{\pi }{4}}{1+\cos \frac{\pi }{4}}}$

$\Rightarrow \tan \frac{\pi }{8}=\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}}$

$\Rightarrow \tan \frac{\pi }{8}=\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$

Rationalising the R.H.S. we get,
$\Rightarrow \tan \frac{\pi }{8}=\sqrt{\frac{(\sqrt{2}-1{)}^2}{(\sqrt{2}+1)(\sqrt{2}-1)}}$

$\Rightarrow \tan \frac{\pi }{8}=\sqrt{\frac{(\sqrt{2}-1{)}^2}{{\sqrt{2}}^2-1^2}}$
$\Rightarrow \tan \frac{\pi }{8}=\sqrt{\frac{(\sqrt{2}-1{)}^2}{2-1}}$

$\Rightarrow \tan \frac{\pi }{8}=\sqrt{2}-1$


c) $\cot \frac{\pi }{8}$
We know that, $\frac{\pi }{8}=\frac{\pi }{2}-\frac{3\pi }{8}$
$\Rightarrow \cot \frac{\pi }{8}=\cot (\frac{\pi }{2}-\frac{3\pi }{8})$
$\Rightarrow \cot \frac{\pi }{8}=\tan \left(\frac{3\pi }{8}\right)$

Also we know that,
$|\cot \theta |=\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}$
Now, put $\theta =\frac{\pi }{8}$

$\therefore \cot \frac{\pi }{8}=\sqrt{\frac{1+\cos 2\frac{\pi }{8}}{1-\cos 2\frac{\pi }{8}}}$

$\Rightarrow \cot \frac{\pi }{8}=\sqrt{\frac{1+\cos \frac{\pi }{4}}{1-\cos \frac{\pi }{4}}}$

$\Rightarrow \cot \frac{\pi }{8}=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}}$

$\Rightarrow \cot \frac{\pi }{8}=\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$

Rationalising the R.H.S. we get,
$\Rightarrow \cot \frac{\pi }{8}=\sqrt{\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}+1)(\sqrt{2}-1)}}$

$\Rightarrow \cot \frac{\pi }{8}=\sqrt{\frac{(\sqrt{2}+1{)}^2}{{\sqrt{2}}^2-1^2}}$
$\Rightarrow \cot \frac{\pi }{8}=\sqrt{\frac{(\sqrt{2}+1{)}^2}{2-1}}$

$\Rightarrow \cot \frac{\pi }{8}=\sqrt{2}+1$