Correct option is: A) 0°
Given that
sin 2A = 2 sin A
= 2 sin A cos A = 2 sin A (\(\because\) sin 2 A = sin A cos A)
= cos A = 1 = cos \(0^\circ\)
= A = \(0^\circ\), Hence, \(\angle\)A = \(0^\circ\)
Alternative Method :
(A) Let \(\angle\)A = \(0^\circ\) then
sin 2 A = sin (2 \(\times\) \(0^\circ\)) = sin \(0^\circ\) = 0
2 sin A = 2 sin \(0^\circ\) = 2 \(\times\) \(0^\circ\) = 0
Hence, sin 2 A = 2 sin A for A = \(0^\circ\)
(B) Let \(\angle\)A = \(30^\circ\)
sin 2A = sin \(60^\circ\) = \(\frac {\sqrt3}{2}\)
2 sin A = 2 sin \(30^\circ\) = 2 \(\times\) \(\frac 12\) = 2 + \(\frac {\sqrt3}{2}\)
(C) Let \(\angle\)A = \(45^\circ\)
sin 2 A = sin \(90^\circ\) = 1
2 sin A = 2 sin \(45^\circ\) = 2 \(\times\) \(\frac 1{\sqrt2}\) = \(\sqrt2\) \(\neq\) 1
(D) Let \(\angle\)A = \(60^\circ\)
sin 2 A = sin \(120^\circ\) = sin (\(90^\circ\) + \(30^\circ\))
= cos \(30^\circ\) = \(\frac {\sqrt3}{2}\)
2 sin A = 2 sin \(60^\circ\) = 2 \(\times\) \(\frac {\sqrt3}{2}\) = \(\sqrt3\) + \(\frac {\sqrt3}{2}\)
