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

Sagot :

xKelvin

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.

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