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?
 A: $$\cos(\alpha-\beta)=\cos\{\theta+\alpha-(\theta+\beta)\}$$
$$=\cos(\theta+\beta)\cos(\theta+\alpha)+\sin(\theta+\beta)\sin(\theta+\alpha)$$
$$\implies\cos(\alpha-\beta)=\cos(\theta+\beta)\cos(\theta+\alpha)+ab$$
$$\implies\{\cos(\alpha-\beta)-ab\}^2=\cos^2(\theta+\beta)\cos^2(\theta+\alpha)$$
Now $\cos^2(\theta+\beta)=1-b^2, \cos^2(\theta+\alpha)=?$
Finally, $\cos2A=2\cos^2A-1$ 
Can you take it from here?
A: 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.
A: Set $\alpha-\beta=\gamma$, $\varphi=\theta+\alpha$, $\psi=\theta+\beta$; then $\gamma=(\theta+\alpha)-(\theta+\beta)=\varphi-\psi$, so
$$
\cos\gamma=
\cos\varphi\cos\psi+
\sin\varphi\sin\psi=
ab+\cos\varphi\cos\psi
$$
Also
$$
\cos2\gamma=2\cos^2\gamma-1=
2(a^2b^2+2ab\cos\varphi\cos\psi+\cos^2\varphi\cos^2\psi)-1
$$
so that
$$
\cos2\gamma-4ab\cos\gamma=
-2a^2b^2+2\cos^2\varphi\cos^2\psi-1
$$
Now $\cos^2\varphi=1-a^2$ and $\cos^2\psi=1-b^2$.
