Question

Simplify :
${\cos }^{-1}(\frac{3}{5}\cos x+\frac{4}{5}\sin x)$

Answer 1

${\cos }^{-1}\left(\frac{3}{5}\cos x+\frac{4}{5}\sin x\right)$

$Let\frac{3}{5}=r\cos \theta \&\frac{4}{5}=r\sin \theta$

then $r^2=\frac{9}{25}+\frac{16}{25}=1$

$=r=1$

$\tan \theta =\frac{4}{\frac{5}{\frac{3}{5}}}=\frac{4}{3}$

$\theta ={\tan }^{-1}\frac{4}{3}$

$\frac{3}{5}\cos x+\frac{4}{5}\sin x$

$=r\cos \theta \cos x+r\sin \theta \sin x$

$=r\left(\cos \left(\theta -x\right)\right)$

$=\cos \left(\theta -x\right)$

therefore ${\cos }^{-1}\left(\frac{3}{5}\cos x+\frac{4}{5}\sin x\right)$

$={\cos }^{-1}\left(\cos \left(\theta -x\right)\right)$

$\theta -x$

$={\tan }^{-1}\frac{4}{3}-x$