Simplify a complicated trigonometric expression

Question

My question is to simplify the following expression:$$\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right)$$

I have this complicated trigonometric expression as above, it is obtained in a geometry problem but I don't know what's the best way to simplify it. The answer for this is $(1 + \sqrt{2})\sin x$ according to WolframAlpha.

My attempt

\begin{align} &\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right)\\ &=\frac{1}{2\sin x}\left(\cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right)\right) \\ &=\frac{1}{2\sin x}\left(1 + \sqrt{2} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos \left(\pi - 2x\right) + \cos\left(\frac{5\pi}{4} - 2x\right)\right) \end{align} $$$$ How to compute after this? Should I use the compound angle formula to expand the $\cos(a + b)$ terms? It seems a bit heavy to me cuz I don't see how it will cancel nicely with $\sin x$. Or should I use the sum to product formula for cos? Thanks in advance!

Answer 1

You are almost there.

$\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right) = \frac{1}{2\sin x}\left(\sin 2x + 1 + \sqrt2 + \cos \left(2\pi - (\frac{3\pi}{4} - 2x\right)\big) - \cos \left(2x\right) + \cos\left(\frac{5\pi}{4} - 2x\right) - \sin2x\right)$

$ = \frac{1}{2\sin x} \ (1 + \sqrt2 + \cos (\frac{5\pi}{4} + 2x) + \cos(\frac{5\pi}{4} - 2x) - \cos(2x) )$

Using $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$,

$ = \frac{1}{2\sin x} \ (1 + \sqrt2 - \sqrt2 \cos (2x) - \cos(2x) )$

$ = \frac{1+\sqrt2}{2\sin x} \ (1 - \cos(2x)) = = \frac{1+\sqrt2}{2\sin x} \ (2 \sin^2x) = (1+\sqrt2) \sin x$

Answer 2

At least, simplify as (for example)

$$\cos \left( \frac{3\pi}{4} - 2x\right)=-\cos \left(2 x+\frac{\pi }{4}\right)$$ $$\cos \left(\pi - 2x\right)=-\cos (2 x)$$ $$ \cos\left(\frac{5\pi}{4} - 2x\right)=-\cos \left(2x-\frac{\pi }{4}\right)$$ $$\cos \left(2 x+\frac{\pi }{4}\right)+\cos \left(2 x-\frac{\pi }{4}\right)=\sqrt{2} \cos (2 x)$$