Question

Prove that
$\sin ({75}^o+\alpha )\sin ({75}^o-\alpha )-\cos ({15}^o+\alpha )\cos ({15}^o-\alpha )=0$

Answer 1

L.H.S$=\sin ({75}^o+\alpha )\sin ({75}^o-\alpha )-\cos ({15}^o+\alpha )\cos ({15}^o-\alpha )$

$=\frac{\cos \left\{\right({75}^o+\alpha )-({75}^o-\alpha \left)\right\}-\cos \left\{\right({75}^o+\alpha )+({75}^o-\alpha \left)\right\}}{2}-\frac{\cos \left\{\right({15}^o+\alpha )+({15}^o-\alpha \left)\right\}+\cos \left\{\right({15}^o+\alpha )-({15}^o-\alpha \left)\right\}}{2}$

$\because \left[\begin{array}{c}2\sin \alpha \sin \beta =\cos (\alpha -\beta )-\cos (\alpha +\beta )\\ 2\cos \alpha \cos \beta =\cos (\alpha +\beta )-\cos (\alpha -\beta )\end{array}\right]$

$=\frac{\cos \left(2\alpha \right)-\cos \left({150}^o\right)}{2}-\frac{\cos {30}^o+\cos \left(2\alpha \right)}{2}$

$=\frac{\cos \left(2\alpha \right)-\cos \left({150}^o\right)-\cos {30}^o-\cos \left(2\alpha \right)}{2}$

$=-\frac{\cos \left({150}^o\right)-\cos {30}^o}{2}$

$-\frac{\cos ({90}^o+{60}^o)-\cos {30}^o}{2}$

$=-\frac{\{-\sin {60}^o\}-\cos {30}^o}{2}$

$=\frac{\sin {60}^o-\cos {30}^o}{2}$

$=\frac{\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}}{2}$

$=0$. R.H.S.