sin(15^circ)^2 %2B sin(75^circ)^2 – InterviewSolution

1.

sin(15^circ)^2 %2B sin(75^circ)^2

Answer»

sin^2(90-75)+sin^2 75cos^2 15+sin^2 151

To find:sin²15°+sin²75°

sin²(90°-75°)+sin²75°=cos²75°+sin²75°=1

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