Question

If $\sin (\theta +\alpha )=a$ and $\sin (\theta +\beta )=b$, prove that $\cos 2(\alpha -\beta )-4ab\cos (\alpha -\beta )=1-2a^2-2b^2$.

Answer 1

$\sin (\theta +\alpha )=a$

$\sin (\theta +\beta )=b$

$\cos (\theta +\alpha )=\sqrt{1-a^2}$

& $\cos (\theta +\beta )=\sqrt{1-b^2}$

$⟶\cos (\alpha -\beta )=\cos \left[\right(\theta +\alpha )-(\theta +\beta \left)\right]$

$=\cos (\theta +\beta )\cos (\theta +\alpha )+\sin (\theta +\beta )\left(\sin \right(\theta +\beta \left)\right)$

$=\left(\sqrt{1-a^2}\right)\left(\sqrt{1-b^2}\right)+ab$

$=ab+\sqrt{1-a^2-b^2+(ab{)}^2}------\left(1\right)$

LHS $\Rightarrow$

$\cos \left(2\right)(\alpha -\beta )-4ab\cos (\alpha -\beta )$

$\Rightarrow 2{\cos }^2(\alpha -\beta )-1-4ab\cos (\alpha -\beta )$

$=2\cos (\alpha -\beta )\left[\cos \right(\alpha -\beta )-2ab]-1$

$\Rightarrow 2(ab+\sqrt{1-a^2-b^2+(ab{)}^2})[ab+\sqrt{1-a^2-b^2+a^2{b}^2}-2ab)-1$

$\Rightarrow 2\left(\sqrt{1-a^2-b^2+a^2{b}^2+ab}\right)\left[\right(\sqrt{1-a^2-b^2+a^2{b}^2})-ab]-1$

$\Rightarrow 2(1-a^2-b^2+a^2{b}^2-a^2{b}^2)-1$

$=2-2a^2-2b^2-1$

$\Rightarrow 1-2a^2-2b^2$

= R.H.S

= Hence Proved.