Question

The value of $4sin\alpha sin\left(\alpha +\frac{\pi }{3}\right)sin\left(\alpha +\frac{2\pi }{3}\right)=$

• A. $sin3\alpha$
• B. $sin2\alpha$
• C. $sin\alpha$
• D. $si{n}^2\alpha$

Answer 1

Answer: (A)

We have, $4sin\alpha sin\left(\alpha +\frac{\pi }{3}\right)sin\left(\alpha +\frac{2\pi }{3}\right)$

$=2sin\alpha \left\{2sin\left(\alpha +\frac{2\pi }{3}\right)sin\left(\alpha +\frac{\pi }{3}\right)\right\}$

$=2sin\alpha \left[2sin\left(\alpha +120^{\circ }\right)sin\left(\alpha +60^{\circ }\right)\right]$
$=2sin\alpha [cos\left(\alpha +120^{\circ }-\alpha -60^{\circ }\right)-$
$cos(\alpha +120^{\circ }+\alpha +60^{\circ })]$
$=2sina[cos60^{\circ }-cos(180^{\circ }+2\alpha )]$
$=2sin\alpha \cdot \frac{1}{2}-2sin\alpha \left(-cos2\alpha \right)$
$=sin\alpha +2cos2\alpha sin\alpha =sin\alpha +sin\left(2\alpha +\alpha \right)-$
$sin\left(2\alpha -\alpha \right)$
$=sin\alpha +sin3\alpha -sin\alpha =sin3\alpha$