<s>[INST] If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$ If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$.
My Attempt:
.$$\sin (\theta+\alpha)=a$$
$$\sin \theta. \cos \alpha+\cos \theta.\sin \alpha=a$$
Multiplying both sides by $2$
$$2\sin \theta.\cos \alpha + 2\cos \theta.\sin \alpha=2a$$
Squaring both sides,
$$4\sin^2 \theta.\cos^2 \alpha + 6\sin^2 \theta. \cos^2 \alpha + 4\cos^2 \theta.\sin^2 \alpha=4a^2$$.
How should I do further?
 [/INST] Consider a circle with radius $\frac{1}{2}$ and a triangle $ABC$ inscribed in the circle with $\angle A = 180^\circ - (\alpha+\theta), \angle B = \theta + \beta$. Then $\angle C = \alpha - \beta$. With the usual notations, we have $a = \sin(\theta+\alpha), b = \sin(\theta+\beta), c= \sin(\alpha - \beta)$. Applying the Cosine rule, $
c^2 = a^2 + b^2 - 2ab \cos C $ gives the desired result.
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