Question

If A + B + C = 180°, then prove that
sin 2A-sin 2B+sin 2C=4 cos A sin B cos C​

Answer 1

Step-by-step explanation:

1. Given A + B + C = 180°
2. 2A + 2B + 2C = 360°
3. 2A + 2B = 360° – 2C
4. sin(2A + 2B) = sin(360° – 20) = – sin2C
5. cos(2A + 2B) = cos(360° – 2C) = cos 2C
6. L.H.S = sin2A – sin2B + sin2C
7. = 2cos(A + B) · sin(A – B) + 2sinC. cosC
8. = – 2cosC.sin(A – B) + 2 sinc.cosC
9. = 2 cosC [sinC – sin (A – B)]
10. = 2 cosC [sin(A + B) – sin(A – B)]
11. = 2 cos C [2cosA . sinB] =
12. 4 cosA sinB.cosC = R.H.S

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