Precalc Trig Identity, verify: $1 + \cos(x) + \cos(2x) = \frac 12 + \frac{\sin(5x/2)}{2\sin(x/2)}$ Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2\cos(3x/2)\cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum. 

Start with the RHS. Notice that 
$$ \sin(5x/2) = \sin(2x + x/2) = \sin 2x \cos (x/2) + \cos 2x \sin (x/2) $$
Also, notice that 
$$ \sin 2x = 2 \sin x \cos x = 4 \sin (x/2) \cos (x/2) \cos x $$
Therefore,
$$ \frac{ \sin (5x/2) }{2 \sin(x/2) } = \frac{4 \sin (x/2) \cos^2 (x/2) \cos x  + \cos 2x \sin (x/2)}{2 \sin(x/2)} $$
$$ = 2 \cos^2 (x/2) \cos x + \frac{ \cos 2x }{2} $$
Also, we have that $\cos^2 (x/2) = \frac{ \cos x + 1 }{2}$ and so
$$ = \cos^2 x + \cos x + \frac{ \cos 2x }{2} $$
$$ = \frac{1+\cos 2x }{2} + \cos x + \frac{ \cos 2x }{2} $$
$$ \cos x + \cos 2x + \frac{1}{2} $$$
add the missing $1/2$ from the RHS and you have the LHS
