Question

sin3x+sin2x-sinx=4Sinx.cosx÷2.cos3x÷2

Answer 1

Answer:

$sinx - siny = 2cos \frac{x + y}{2} sin \frac{x - y}{2} \\ sin2x = 2sinx \: cosx \\ cosx + cosy = 2cos \frac{x + y}{2} cos \frac{x - y}{2}$

Solution

$sin3x - sinx \: + \: sin2x \\ = 2cos2x \: sinx \: + 2sinx \: cosx \\ = 2 \: sinx \: (cos2x + cosx) \\ = 2 \: sinx \: (2cos \frac{3x}{2} cos \frac{x}{2} ) \\ = 4sinx \: cos \frac{3x}{2} cos \frac{x}{2}$

Hence proved

Answer 2

Answer:

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