Question

Prove that :

$\cos {18}^o-\sin {18}^o=\sqrt{2}\sin {27}^o$.

Answer 1

$L.H.S\Rightarrow$

$\cos {18}^o-\sin {18}^o=\cos {18}^o-\cos {72}^o$

$=2\sin \left(\frac{{18}^o+{72}^o}{2}\right)\sin \left(\frac{{72}^o-{18}^o}{2}\right)$

$=2\sin 45\sin {27}^o$

$=\sqrt{2}\sin {27}^o=R.H.S$