Prove Prove sin⁡5A−sin⁡A+sin⁡2A=sin⁡2A(2cos⁡3A+1)

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Prove Prove sin⁡5A−sin⁡A+sin⁡2A=sin⁡2A(2cos⁡3A+1)

Obiajulu · Accepted Answer

Notice that if you distribute the sin2A on the right side of $\sin 5 A - \sin A + \sin 2 A = \sin 2 A (2 \cos 3 A + 1)$ you get $\sin 5 A - \sin A + \sin 2 A = 2 \sin 2 A \cos 3 A + \sin 2 A$ . Since both sides have $\sin 2 A$ , this means we need to rewrite $\sin 5 A - \sin A = 2 \sin 2 A \cos 3 A .$
The Sum-To-Product Formula for a difference of sines is $\sin u - \sin v = 2 \cos (\frac{u + v}{2}) \sin ((\frac{u - v}{2}) .$
Therefore, if $\sin u - \sin v = \sin 5 A - \sin A, t h e n u = 5 A and v = A$ . Using the Sum-to-Product Formula then gives:
$\sin 5 A - \sin A + \sin 2 A$
$= 2 \cos \frac{5 A + A}{2} \sin \frac{5 A - A}{2} + \sin 2 A$ Use the Sum-to-Product formula
$= 2 \cos \frac{6 A}{2} \sin \frac{4 A}{2} + \sin 2 A$ Simplify.
$= 2 \cos 3 A \sin 2 A + \sin 2 A$ Simplify.
$= \sin 2 A (2 \cos 3 A + 1)$ Factor out $\sin 2 A$ .

Prove Prove sin5A−sin A+sin2A=sin2A(2cos3A+1)

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