Question

Prove that:
sin3A + sin2A - sinA = 4sinA × cosA/2 × cos 3A/2

Answer 1

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Answer 2

Make use of the following trigonometric identities.

$\sin 2\theta=2\sin \theta\cos \theta$

Here make use of the sum formula,

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

Thus,

$\sin 2A-\sin A=2\cos (\frac{3A}{2}) \sin (\frac{A}{2})$.

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

The proof is complete.