Answer: see proof below
Step-by-step explanation:
Use the following Power Reducing Identities:
[tex]\cos^2 A = \dfrac{1}{2}(1 + \cos 2A)\\\\\sin^2 A = \dfrac{1}{2}(1 - \cos 2A)[/tex]
Use the following Product to Sum Identity:
[tex]\cos A\cdot \cos B=\dfrac{1}{2}\bigg[\cos(A+B)+\cos(A-B)\bigg][/tex]
Proof LHS → RHS
[tex]\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[/tex]
[tex]\text{Power Reducing:}\qquad \dfrac{1}{2}\bigg(1+\cos 2\beta\bigg)\dfrac{1}{2}\bigg(1-\cos 2\beta\bigg)\dfrac{1}{2}\bigg(1-\cos 2\beta\bigg)\\\\\\.\qquad \qquad \qquad \quad =\dfrac{1}{8}(1+\cos 2\beta)(1-\cos 2\beta)(1-\cos 2\beta)[/tex]
[tex]\text{Expand:}\qquad \qquad \dfrac{1}{8}\bigg(1-\cos 2\beta-\cos^2 2\beta+\cos^3 2\beta\bigg)\\\\.\qquad \qquad \qquad =\dfrac{1}{8}\bigg(1-\cos 2\beta-\cos^2 2\beta+\cos^2 2\beta \cdot \cos \2\beta\bigg)[/tex]
[tex]\text{Power Reducing:}\qquad \dfrac{1}{8}\bigg(1-\cos 2\beta -\dfrac{1}{2}(1+\cos 4\beta)+\dfrac{1}{2}(1+\cos 4\beta)\cos 2\beta\bigg)\\\\.\qquad \qquad \qquad =\dfrac{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 =\dfrac{1}{16}\bigg(1-\cos 2\beta -\cos 4\beta+\cos 4\beta \cdot \cos 2\beta \bigg)[/tex]
[tex]\text{Product to Sum:}\quad \dfrac{1}{16}\bigg(1-\cos 2\beta -\cos 4\beta +\dfrac{1}{2}[\cos(4\beta+2\beta)+\cos (4\beta-2\beta)]\bigg)\\\\.\qquad \qquad \qquad =\dfrac{1}{32}\bigg(2-2\cos 2\beta -2\cos 4\beta +\cos 6\beta +\cos 2\beta\bigg)\\\\.\qquad \qquad \qquad =\dfrac{1}{32}\bigg(\cos 6\beta-2\cos 4\beta -\cos 2\beta +2 \bigg)[/tex]
[tex]\text{LHS=RHS:}\\ \dfrac{1}{32}\bigg(\cos 6\beta-2\cos 4\beta -\cos 2\beta +2 \bigg)=\dfrac{1}{32}\bigg(\cos 6\beta-2\cos 4\beta -\cos 2\beta +2 \bigg)[/tex]
