SIN2A-SIN2B-SIN2C=4COSA SINB COSC
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Answer:
If A + B + C = 180°. Prove that: sin2A – sin2B + sin2C = 4cosA ·
Step-by-step explanation:
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.
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