Question

sin 3 theta into cos cube theta + cos 3 theta into sin cube theta

Answer 1

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$= \sin(3 \theta) \cos ^{3} (\theta) + \cos(3 \theta) \sin^{3} (\theta) \\ \\ = (3 \sin( \theta) - 4 \sin ^{3} (\theta) )cos ^{3} (\theta) + \\ (4\cos ^{3}(\theta) - 3\cos( \theta) ) sin ^{3} (\theta) \\ \\ = 3\sin( \theta)cos ^{3} (\theta) - 4sin ^{3} (\theta)cos ^{3} (\theta) + \\ 4sin ^{3} (\theta)cos ^{3} (\theta) - 3\cos( \theta)sin ^{3} (\theta) \\ \\ = 3\sin( \theta)cos ^{3} (\theta) - 3\cos( \theta)sin ^{3} (\theta) \\ \\ = 3\cos( \theta)sin (\theta)(cos ^{2} (\theta) - sin ^{2} (\theta)) \\ \\ = \frac{3}{2} \sin( 2\theta)\cos( 2\theta) \\ \\ = \frac{3}{4} \sin( 4\theta)$

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Answer 2

Answer:

Step-by-step explanation:

Answer is given in the image