Question

prove that :cos18-sin18=√2 sin27
NOTE_>√2 IS SEPERATE

Answer 1

I think it help you .See the picture

Answer 2

Answer:

To prove : cos 18° - sin 18° = √2 sin 27°

L.H.S.

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

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

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

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

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

$= 2 cos \frac{90^{\circ}}{2}. sin \frac{54^{\circ}}{2}$

$=2 cos 45^{\circ} . sin 27^{\circ}$  

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

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

= R.H.S.

Hence, proved.