(SOLVED) Simplify $\cos^{-1}\left( \frac{7}{2}\left( 1+\cos{2x} \right) + \sqrt{\left( \sin^{2}{x} - 48\cos^{2}{x} \right)}\sin{x}\right)$

Question

I need a little help in simplification of an ITF problem of a college entrance exam I got stuck in.

$$\cos^{-1}\left( \frac{7}{2}\left( 1+\cos{2x} \right) + \sqrt{\left( \sin^{2}{x} - 48\cos^{2}{x} \right)}\sin{x}\right)$$

with
$ x\in \left( 0,\frac{\pi}{2} \right) $

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What I'm doing:

$y=\cos ^{-1}\left(\frac{7}{2}(1+\cos 2 x)+\sqrt{\left(\sin ^2 x-48 \cos ^2 x\right)} \sin x\right)$

$\quad=\cos ^{-1}\left((7 \cos x)(\cos x)+\sqrt{1-49 \cos ^2 x} \sqrt{1-\cos ^2 x}\right)$ ......stuck

Thanks!

Answer 1

Suppose $\cos^{-1}\frac17\le x\le\frac\pi2$, and take $\theta$ in $[0,\frac\pi2]$ such that $\cos\theta=7\cos x$. Then $$\eqalign{ \frac{7}{2}(1+\cos 2 x)+{}&\sqrt{\left(\sin ^2 x-48 \cos ^2 x\right)} \sin x\cr &=(7 \cos x)(\cos x)+\sqrt{1-49 \cos ^2 x}\sin x\cr &=\cos x\cos\theta+\sin x\sin\theta\cr &=\cos(x-\theta)\cr}$$ and so $$\cos^{-1}({\rm this})=x-\theta=x-\cos^{-1}(7\cos x)\ ,$$ provided $0\le x-\cos^{-1}(7\cos x)\le\pi$, and this is true for all $x$ in the range stated above.