Question

Find the simplified form of $co{s}^{-1}(\frac{3}{5}cosx+\frac{4}{5}sinx)wherex\in [\frac{-3\pi }{4},\frac{\pi }{4}].$

Answer 1

We have, $co{s}^{-1}(\frac{3}{5}cosx+\frac{4}{5}sinx)wherex\in [\frac{-3\pi }{4},\frac{\pi }{4}]$

Let $cosy=\frac{3}{5}$
$\Rightarrow siny=\frac{4}{5}\Rightarrow y=co{s}^{-1}\frac{3}{5}=si{n}^{-1}\frac{4}{5}=ta{n}^{-1}\left(\frac{4}{3}\right)\therefore co{s}^{-1}[cosy.cosx+siny.sinx]=co{s}^{-1}\left[cos\right(y-x\left)\right][\because cos(A-B)=cosA.cosB+sinA.sinB]=y-x=ta{n}^{-1}\frac{4}{3}-x[\because y=ta{n}^{-1}\frac{4}{3}]$