Question

prove that Sin2A+ sin2B+ sin2C =4cosA cosB cosC​

Answer 1

Answer:

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Answer 2

Given A + B + C = 180° 2A + 2B + 2C = 360° 2A + 2B = 360° – 2C sin(2A + 2B) = sin(360° – 20) = – sin2C cos(2A + 2B) = cos(360° – 2C) = cos 2C L.H.S = sin2A – sin2B + sin2C = 2cos(A + B) · sin(A – B) + 2sinC. cosC = – 2cosC.sin(A – B) + 2 sinc.cosC = 2 cosC [sinC – sin (A – B)] = 2 cosC [sin(A + B) – sin(A – B)] = 2 cos C [2cosA . sinB] = 4 cosA sinB.cosC = R.H.S.Read more on Sarthaks.com - https://www.sarthaks.com/643786/if-a-b-c-180-prove-that-sin2a-sin2b-sin2c-4cosa-sinb-cosc