Simplify expression $\frac{2\cos(x)+1}{4\cos(x/2+π/6)}$ How to simplify the following expression: $$\frac{2\cos(x)+1}{4\cos\left(\frac x2+\fracπ6\right)}$$
I got to: $ \dfrac{2\cos(x)+1}{4\cos\left(\dfrac x2\right)\cdot \dfrac{\sqrt3}2-\sin(x) \cdot \frac 12}$
 A: Let $\dfrac{x}{2}+\dfrac{\pi}{6}=y$. Then $x=2y-\dfrac{\pi}{3}$.
\begin{align*}
\frac{2\cos x+1}{4\cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)}&=\frac{2\cos\left(2y-\dfrac{\pi}{3}\right)+1}{4\cos y}\\
&=\frac{2\cos2y\cos\dfrac{\pi}{3}+2\sin2y\sin\dfrac{\pi}{3}+1}{4\cos y}\\
&=\frac{2\cos^2y-1+2\sqrt{3}\sin y\cos y+1}{4\cos y}\\
&=\frac{\cos y+\sqrt{3}\sin y}{2}\\
&=\cos y\cos\frac{\pi}{3}+\sin y\sin\frac{\pi}{3}\\
&=\cos\left(y-\frac{\pi}{3}\right)\\
&=\cos\left(\frac{x}{2}-\frac{\pi}{6}\right)
\end{align*}
A: The denominator can be simplified to $$4(\cos(x/2)\cos(\pi/6)-\sin(x/2)\sin(\pi/6))=2(\sqrt3\cos(x/2)-\sin(x/2))$$
The numerator is $$2\cos(x)+1=2(2\cos^2(x/2)-1)+1=4\cos^2(x/2)-1$$
So you have $$\require{cancel}\frac{4\cos^2(x/2)-1}{2\sqrt3\cos(x/2)-2\sin(x/2)}=\frac{4\cos^2(x/2)-\cos^2(x/2)-\sin^2(x/2)}{2\sqrt3\cos(x/2)-2\sin(x/2)}\\=\frac{3\cos^2(x/2)-\sin^2(x/2)}{2\sqrt3\cos(x/2)-2\sin(x/2)}\\=\frac{(\sqrt3\cos(x/2)-\sin(x/2))(\sqrt3\cos(x/2)+\sin(x/2))}{2(\sqrt3\cos(x/2)-\sin(x/2))}\\=\frac{\cancel{(\sqrt3\cos(x/2)-\sin(x/2)}(\sqrt3\cos(x/2)+\sin(x/2))}{2\cancel{(\sqrt3\cos(x/2)-\sin(x/2))}}\\=\frac12(\sqrt3\cos(x/2)+\sin(x/2))$$
Edit:
As pointed out in the comments, this can be simplified further
$$=\left(\frac{\sqrt3}2\cos(x/2)+\frac12\sin(x/2)\right)=\cos(\pi/6)\cos(x/2)+\sin(x/2)\sin(\pi/6)\\=\cos\left(\frac x2-\frac\pi6\right)$$
Or equivalently, $\sin\left(\frac\pi3+\frac x2\right)$
A: Hint:
Take out $2$ as common factor in the numerator
Now using 
Prosthaphaeresis Formula, $\cos x+\cos\dfrac\pi3=2\cos\left(\dfrac x2+\dfrac\pi6\right)\cos\left(\dfrac x2-\dfrac\pi6\right)$
A: for the denominator you Can write $$\cos(x)=2\cos^2(x/2)-1$$
and your denominator must be $$2\sqrt{3}\cos(x/2)-2\sin(x/2)$$
A: Using half-angle formula: $\cos\dfrac A2=\pm\sqrt{\dfrac{1+\cos A}2}$ with $A=x+\dfrac\pi3$ to give $$\cos\left(\frac x2 + \frac \pi6 \right)=\pm\sqrt{\dfrac{1+\cos \left(x + \frac \pi3 \right)}2}=\pm\sqrt{\dfrac{1+\frac{\sqrt3}2\cos x-\frac12 \sin x }2}$$ so $$\cos\left(\frac x2 + \frac \pi6 \right)=\pm\frac{\sqrt{1+\sqrt3\cos x-\sin x}}2.$$ Hence $$\frac{2 \cos x + 1}{4\cos\left(\frac x2 + \frac \pi6 \right)}=\pm \frac{2(2 \cos x + 1)}{\sqrt{1+\sqrt3\cos x-\sin x}}$$ which is positive if $\dfrac x2 + \dfrac \pi6$ is in quadrant I or IV and negative if $\dfrac x2 + \dfrac \pi6$ is in quadrant II or III.
