Question

prove that cos 18 - sin 18 = 2 sin 27

Answer 1

Cos18- Sin18= cos18 - cos72= -2×sin(18+72)/2 × sin (18-72)/2 (using CosC-CosD) = -2 sin45×sin(-27) 2sin45sin27 (sin(-x)=-sinx) =√2×sin27

Answer 2

Answer:

We have to prove :  cos 18° - sin 18° = √2 sin 27°

L.H.S.

$cos 18^{\circ} - sin 18^{\circ}$

$= cos 18^{\circ} - sin (90 - 72)^{\circ}$

$= cos 18^{\circ} - cos 72^{\circ}$               ( Because, cos x = sin (90 - x ) )

$=2 sin\frac{18+72}{2} . sin \frac{72-18}{2}$

( Because, cos A - cos B = 2 sin $\frac{A+B}{2}$ . sin $\frac{B-A}{2}$ )

$=2 sin \frac{90}{2} . sin \frac{54}{2}$

$=2\times \frac{1}{\sqrt{2}} . sin 27^{\circ}$  ( sin 45° = 1/√2 )

$=\sqrt{2} . sin 27^{\circ}$

R.H.S.

Hence, proved.