Question

Prove that:
Sin^2A+Sin^2B.Cos2A=Sin^2B+Sin^2A.Cos2B​

Answer 1

See Below.

Explanation:

We want to prove that:

`\displaystyle \sin^2 A + \sin^2 B \cdot \cos 2A = \sin^2 B + \sin^2 A \cdot \cos 2B`

Recall that double-angle identity for cosine:

`\displaystyle \begin{aligned} \cos 2x &= \cos^2x - \sin^2 x \\ &= 2\cos^2x -1 \\ &= 1 - 2\sin^2 x\end{aligned}`

Substitute cos(2A) for its third form:

`\displaystyle \sin^2 A + \sin^2 B \cdot \left(1 - 2\sin^2 A\right) = \sin^2 B + \sin^2 A \cdot \cos 2B`

Distribute:

`\displaystyle \sin^2 A + \sin^2 B - 2\sin^2B \sin^2A = \sin^2 B + \sin^2 A \cdot \cos 2B`

Rewrite:

`\displaystyle \sin^2 B + \left(\sin^2 A - 2\sin^2 B\sin^2 A\right)`

Factor:

`\displaystyle \sin^2 B + \sin^2A\left(1 - 2\sin^2 B\right) = \sin^2 B + \sin^2A\cdot \cos 2B`

Double-Angle Identity for cosine:

`\displaystyle \sin^2 B + \sin^2 A \cdot \cos 2B \stackrel{\checkmark}{=} \sin ^2 B + \sin^2 A\cdot \cos 2B`

Hence proven.