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Circular Measurement of Angle

In a triangle...

Question

In a triangle ABC, prove that $sin 2 m A + sin 2 m B + sin 2 m C = (- 1)^{m + 1} 4 sin m A sin m B sin m C$ , where m is any integer.

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Solution

Hint: L.H.S.
$= 2 sin (m A + m B) cos (m A - m B) + 2 sin m C cos m C$
$= 2 sin (m π - m C) cos (m A - m B) + 2 sin m C cos [m π - m (A + B)]$
$= 2 (- 1)^{m - 1} sin m C [cos (m A - m B) - cos (m A + m B)]$
$= 4 (- 1)^{m - 1} sin m A sin m B sin m C$ .

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