Question

If ${\rm{A}} = 60^\circ $ verify the following $\sin 2{\rm{A}} = 2\sin {\rm{A}}\cos {\rm{A}}$

Answer 1

Hint:
Put ${\rm{A}} = 60^\circ $ in L.H.S and R.H.S separately, then prove LHS = RHS

Complete step by step solution:
LHS
$ = \sin 2{\rm{A}}$
$ = \sin 2 \times 60^\circ $
$ = \sin 120^\circ $
$ = \sin \left( {180^\circ - 60^\circ } \right) = \sin 60^\circ $
$ = \sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$
RHS
$ = 2\sin {\rm{A}}\cos {\rm{A}}$
$ = 2\sin 60^\circ \cos 60^\circ $
$ = 2 \times \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{2}$
$ = \dfrac{{\sqrt 3 }}{2}$
So, L.H.S = R.H.S
Hence verified.

Note:
There is no hard and fast rule to prove trigonometric expressions equal. We just use correct formulas at the correct place. We should remember the value of trigonometric ratios at basic angles. For example, in this question the value of sin and cos at 60 degrees is necessary to remember.