First slide

Trigonometric transformations

Question

$\sin^{2} A + \sin^{2} (A - B) + 2 \sin A \cos B \sin (B - A)$ is equal to

Moderate

A

$\sin^{2} A$
B

$\sin^{2} B$
C

$\cos^{2} A$
D

$\cos^{2} B$

Solution

$\begin{array}{l} \sin^{2} A + \sin^{2} (A - B) + [\sin (A + B) + \sin (A - B)] \sin (B - A) \\ = \sin^{2} A + \sin^{2} (A - B) + \sin (A + B) \sin (B - A) - \sin^{2} (A - B) \\ = \sin^{2} A + \sin^{2} B - \sin^{2} A = \sin^{2} B \end{array}$

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