Question

prove :-
sin3A+sin 2A-sinA=4sinA cosA/2 cos3A/2

Answer 1

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please refer to the attachment above

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Answer 2

$LHS = \sin(3A) + \sin(2A) - \sin(A) \\ \\= \sin(3A) - \sin(A) + \sin(2A) \\\\ = 2 \cos( \frac{3A + A}{2} ) \sin( \frac{3A - A}{2} ) +\sin2A\\ \\= 2 \cos( \frac{4A}{2} ) \sin( \frac{2A}{2} ) + \sin2A$

=2 cos2A sinA +2 sinA cosA

=2 sin A [cos2A + cosA]

=2 sin A [2 cos (2A+A)/2 cos (2A-A)/2]

=2 sin A [2 cos 3A/2 cos A/2 ]

=4 sin A cos A/2 cos 3A/2

= RHS