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. 
 A: 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
A: For $\sin\frac{x}{2}\neq0$ we obtain:
$$1+\cos{x}+\cos2x=\frac{2\sin\frac{x}{2}+2\sin\frac{x}{2}\cos{x}+2\sin\frac{x}{2}\cos2x}{2\sin\frac{x}{2}}=$$
$$=\frac{2\sin\frac{x}{2}+\sin\frac{3x}{2}-\sin\frac{x}{2}+\sin\frac{5x}{2}-\sin\frac{3x}{2}}{2\sin\frac{x}{2}}=\frac{1}{2}+\frac{\sin\frac{5x}{2}}{2\sin\frac{x}{2}}.$$
I used the following formula.
$$\sin\alpha\cos\beta=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta)).$$
For example,$$2\sin\frac{x}{2}\cos{x}=2\cdot\frac{1}{2}\left(\sin\left(\frac{x}{2}+x\right)+\sin\left(\frac{x}{2}-x\right)\right)=\sin\frac{3x}{2}-\sin\frac{x}{2}.$$
A: Hint:
Use 
http://mathworld.wolfram.com/WernerFormulas.html
$2\sin\dfrac x2\cos mx=\sin\left(m+\dfrac12\right)x-\sin\left(m-\dfrac12\right)x$
Set $m=1,2$ and add to find $$2\sin\dfrac x2(\cos x+\cos2x)=\sin\dfrac{5x}2-\sin\dfrac x2$$
Assuming $\sin\dfrac x2\ne0,$ divide both sides by $2\sin\dfrac x2$
See also: How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
A: The left-hand side is $$\Re(1+e^{ix}+e^{2ix})=\Re\frac{e^{3ix}-1}{e^{ix}-1}=\Re\frac{2ie^{3ix/2}\sin\frac{3x}{2}}{2ie^{ix/2}\sin\frac{x}{2}}=\Re\frac{2e^{ix}\sin\frac{3x}{2}}{2\sin\frac{x}{2}}\\=\frac{2\cos x\sin\frac{3x}{2}}{2\sin\frac{x}{2}}=\frac{\sin\frac{x}{2}+\sin\frac{5x}{2}}{2\sin\frac{x}{2}}=\frac12+\frac{\sin\frac{5x}{2}}{2\sin\frac{x}{2}}.$$
