Answer:
See Below.
Step-by-step explanation:
We want to prove that:
[tex]\displaystyle \sin^2 A + \sin^2 B \cdot \cos 2A = \sin^2 B + \sin^2 A \cdot \cos 2B[/tex]
Recall that double-angle identity for cosine:
[tex]\displaystyle \begin{aligned} \cos 2x &= \cos^2x - \sin^2 x \\ &= 2\cos^2x -1 \\ &= 1 - 2\sin^2 x\end{aligned}[/tex]
Substitute cos(2A) for its third form:
[tex]\displaystyle \sin^2 A + \sin^2 B \cdot \left(1 - 2\sin^2 A\right) = \sin^2 B + \sin^2 A \cdot \cos 2B[/tex]
Distribute:
[tex]\displaystyle \sin^2 A + \sin^2 B - 2\sin^2B \sin^2A = \sin^2 B + \sin^2 A \cdot \cos 2B[/tex]
Rewrite:
[tex]\displaystyle \sin^2 B + \left(\sin^2 A - 2\sin^2 B\sin^2 A\right)[/tex]
Factor:
[tex]\displaystyle \sin^2 B + \sin^2A\left(1 - 2\sin^2 B\right) = \sin^2 B + \sin^2A\cdot \cos 2B[/tex]
Double-Angle Identity for cosine:
[tex]\displaystyle \sin^2 B + \sin^2 A \cdot \cos 2B \stackrel{\checkmark}{=} \sin ^2 B + \sin^2 A\cdot \cos 2B[/tex]
Hence proven.
