Question

please someone help me to prove this. ​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the following Power Reducing Identities:

$\sin^3\theta= \dfrac{1+\cos 2\theta}{2}\\\\\\\cos^2\theta=\dfrac{3\sin \theta-\sin 3\theta}{4}$

Use the following Product to Sum Identity:

$\sin A \cdot \cos B = \dfrac{1}{2}\bigg[\sin (A + B) + \sin (A - B)\bigg]$

Proof LHS → RHS

LHS:                           cos² Ф · sin³ Ф

$\text{Power Reducing:}\qquad \bigg(\dfrac{1+\cos 2\theta}{2}\bigg)\bigg(\dfrac{3\sin \theta-\sin 3\theta}{4}\bigg)$

$\text{Expand:}\qquad \qquad \dfrac{1}{8}(3\sin \theta -\sin 3\theta +3\sin \theta\cdot cos 2\theta-\sin 3\theta \cdot \cos 2\theta)$

$\text{Product to Sum:}\\ \dfrac{1}{8}\bigg[3\sin \theta -\sin 3\theta +\dfrac{3}{2}\bigg(\sin (\theta +2\theta) +\sin(\theta -2\theta)\bigg)-\dfrac{1}{2}\bigg(\sin (3\theta+2\theta) +\sin (3\theta-2\theta)\bigg)\bigg]$

$\text{Simplify:}\quad \dfrac{1}{8}\bigg[3\sin \theta -\sin 3\theta +\dfrac{3}{2}\bigg(\sin 3\theta +\sin(-\theta)\bigg)-\dfrac{1}{2}\bigg(\sin 5\theta +\sin \theta\bigg)\bigg]$

$\text{Cofunction:}\quad \dfrac{1}{8}\bigg[3\sin \theta -\sin 3\theta +\dfrac{3}{2}\bigg(\sin 3\theta -\sin(\theta)\bigg)-\dfrac{1}{2}\bigg(\sin 5\theta +\sin \theta\bigg)\bigg]$

$\text{Simplify:}\qquad \dfrac{1}{16}\bigg(6\sin \theta -2\sin 3\theta +3\sin 3\theta -3\sin(\theta)-2\sin 5\theta -2\sin \theta\bigg)\\\\\\.\qquad \qquad =\dfrac{1}{16}\bigg(2\sin \theta + \sin 3\theta -\sin 5\theta\bigg)$

$\text{LHS=RHS:}\quad \dfrac{1}{16}\bigg(2\sin \theta + \sin 3\theta -\sin 5\theta\bigg)=\dfrac{1}{16}\bigg(2\sin \theta + \sin 3\theta -\sin 5\theta\bigg)\quad \checkmark$