<s>[INST] 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}$
 [/INST] 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.
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