Question

If $\cos (\theta -\alpha )=a$ and $\sin (\theta -\beta )=b$ $(0<\theta -\alpha ,\theta -\beta <\frac{\pi }{2})$, then prove that ${\cos }^2(\alpha -\beta )+2ab\sin (\alpha -\beta )$ is equal to $a^2-b^2$.

Answer 1

We have $\sin (\alpha -\beta )=\sin (\alpha -\theta +\theta -\beta )=\sin [(\theta -\beta )-(\theta -\alpha )]$
$=\sin (\theta -\beta )\cos (\theta -\alpha )-\cos (\theta -\beta )\sin (\theta -\alpha )=ba-\sqrt{1-b^2}\sqrt{1-a^2}$
and $\cos (\alpha -\beta )=\cos (\alpha -\theta +\theta -\beta )=\cos [(\theta -\beta )-(\theta -\alpha )]$
$=\cos (\theta -\beta )\cos (\theta -\alpha )+\sin (\theta -\beta )\sin (\theta -\alpha )=a\sqrt{1-b^2}+b\sqrt{1-a^2}$
Substituting these values in the given expression, we get
${\cos }^2(\alpha -\beta )+2ab={(a\sqrt{1-b^2}+b\sqrt{1-a^2})}^2+2ab[ba-\sqrt{1-b^2}\sqrt{1-a^2}]=a^2(1-b^2)+b^2(1-a^2)+2ab\sqrt{1-a^2}\sqrt{1-b^2}+2a^2{b}^2-2ab\sqrt{1-a^2}\sqrt{1-b^2}=a^2-b^2$