Question

Please someone help me to prove this. ​

Answer 1

Answer: see proof below

Explanation:

Use the following Power Reducing Identities:

`\sin^3\theta= (1+\cos 2\theta)/(2)\\\\\\\cos^2\theta=(3\sin \theta-\sin 3\theta)/(4)`

Use the following Product to Sum Identity:

`\sin A \cdot \cos B = (1)/(2)\bigg[\sin (A + B) + \sin (A - B)\bigg]`

Proof LHS → RHS

LHS: cos² Ф · sin³ Ф

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

`\text{Expand:}\qquad \qquad (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:}\\ (1)/(8)\bigg[3\sin \theta -\sin 3\theta +(3)/(2)\bigg(\sin (\theta +2\theta) +\sin(\theta -2\theta)\bigg)-(1)/(2)\bigg(\sin (3\theta+2\theta) +\sin (3\theta-2\theta)\bigg)\bigg]`

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

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

`\text{Simplify:}\qquad (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 =(1)/(16)\bigg(2\sin \theta + \sin 3\theta -\sin 5\theta\bigg)`

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