<s>[INST] Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality: 

$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$

I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\left(-4+3 \sin^2 \theta\right)$ from the right side, but I don't know where to go from there.
 [/INST] The extreme values of $A\cos x +B\sin x$ are $\pm\sqrt {A^2+B^2}.$ This is obvious and trivial if $A=B=0.$ If $A,B$ are not both $0,$ there exists $y$ such that $\cos y=A/\sqrt {A^2+B^2}$ and $\sin y=B/\sqrt {A^2+B^2},$ so $$|A\cos x +B\sin x| =|(\sqrt {A^2+B^2}\;)(\cos x \cos y+\sin x \sin y)|$$ $$=(\sqrt {A^2+B^2}\;) |\cos (x-y)|\leq \sqrt {A^2+B^2}.$$
Putting $\cos^2 \theta=(1+\cos 2 \theta)/2,$ the inequality in the Q can be re-arranged as $5\geq -4\sin 2\theta -3\cos 2 \theta .$ Which is true because $|-4\sin 2 \theta -3\cos 2 \theta| \leq \sqrt {(-4)^2+(-3)^2}=5.$
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