Question

$\cos {18}^0-\sin {18}^0=\sqrt{a}\sin {27}^0$ Find a

Answer 1

L.H.S.$=\cos {18}^0-\sin {18}^0=\cos {18}^0-\sin ({90}^0-{72}^0)=\cos {18}^0-\cos {72}^0$
$=2\sin \frac{{18}^0+{72}^0}{2}\sin \frac{{72}^0-{18}^0}{2}$
$=2\sin {45}^0\sin {27}^0=2\frac{1}{\sqrt{2}}\sin {27}^0=\sqrt{2}\sin {27}^0$
Ans: 2