Question

Please someone help me to prove this...​

Answer 1

Answer: see proof below

Explanation:

Use the following Power Reducing Identities:

`\cos^2 A = (1)/(2)(1 + \cos 2A)\\\\\sin^2 A = (1)/(2)(1 - \cos 2A)`

Use the following Product to Sum Identity:

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

Proof LHS → RHS

`\text{LHS:}\qquad \qquad \qquad \cos^2\beta\cdot \sin^4\beta\\.\qquad \qquad \qquad =\cos^2\beta\cdot \sin^2\beta\cdot \sin^2 \beta`

`\text{Power Reducing:}\qquad (1)/(2)\bigg(1+\cos 2\beta\bigg)(1)/(2)\bigg(1-\cos 2\beta\bigg)(1)/(2)\bigg(1-\cos 2\beta\bigg)\\\\\\.\qquad \qquad \qquad \quad =(1)/(8)(1+\cos 2\beta)(1-\cos 2\beta)(1-\cos 2\beta)`

`\text{Expand:}\qquad \qquad (1)/(8)\bigg(1-\cos 2\beta-\cos^2 2\beta+\cos^3 2\beta\bigg)\\\\.\qquad \qquad \qquad =(1)/(8)\bigg(1-\cos 2\beta-\cos^2 2\beta+\cos^2 2\beta \cdot \cos \2\beta\bigg)`

`\text{Power Reducing:}\qquad (1)/(8)\bigg(1-\cos 2\beta -(1)/(2)(1+\cos 4\beta)+(1)/(2)(1+\cos 4\beta)\cos 2\beta\bigg)\\\\.\qquad \qquad \qquad =(1)/(16)\bigg(2-2\cos 2\beta -1-\cos 4\beta +\cos 2\beta +\cos 4\beta \cdot \cos 2\beta \bigg)\\\\.\qquad \qquad \qquad =(1)/(16)\bigg(1-\cos 2\beta -\cos 4\beta+\cos 4\beta \cdot \cos 2\beta \bigg)`

`\text{Product to Sum:}\quad (1)/(16)\bigg(1-\cos 2\beta -\cos 4\beta +(1)/(2)[\cos(4\beta+2\beta)+\cos (4\beta-2\beta)]\bigg)\\\\.\qquad \qquad \qquad =(1)/(32)\bigg(2-2\cos 2\beta -2\cos 4\beta +\cos 6\beta +\cos 2\beta\bigg)\\\\.\qquad \qquad \qquad =(1)/(32)\bigg(\cos 6\beta-2\cos 4\beta -\cos 2\beta +2 \bigg)`

`\text{LHS=RHS:}\\ (1)/(32)\bigg(\cos 6\beta-2\cos 4\beta -\cos 2\beta +2 \bigg)=(1)/(32)\bigg(\cos 6\beta-2\cos 4\beta -\cos 2\beta +2 \bigg)`