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  4. cos -1 (1/√2)( cos (7 π/5)- sin (2 π/5)) is equal to

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Q. $\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{7 \pi}{5}-\sin \frac{2 \pi}{5}\right)\right\}$ is equal to

Inverse Trigonometric Functions

Solution:

$\frac{1}{\sqrt{2}}\left(\cos \frac{7 \pi}{5}-\sin \frac{2 \pi}{5}\right)=\cos \frac{\pi}{4} \cos \frac{7 \pi}{5}-\sin \frac{\pi}{4} \sin \frac{2 \pi}{5}=\cos \frac{\pi}{4} \cos \frac{7 \pi}{5}+\sin \frac{\pi}{4} \sin \frac{7 \pi}{5}$
$=\cos \left(\frac{7 \pi}{5}-\frac{\pi}{4}\right)=\cos \left(\frac{23 \pi}{20}\right)=\cos \left(\pi+\frac{3 \pi}{20}\right)=\cos \left(\pi-\frac{3 \pi}{20}\right)=\cos \frac{17 \pi}{20}$
$\therefore \cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{7 \pi}{5}-\sin \frac{2 \pi}{5}\right)\right\}=\cos ^{-1}\left(\cos \left(\frac{17 \pi}{20}\right)\right)=\frac{17 \pi}{20}\left(\right.$ since $\frac{17 \pi}{20}$ lies between 0 and $\left.\pi\right)$