Question

a+b+c=90 then prove sin2a+ sin2b +sin2c=4 cosacosbcosc

Answer 1

 
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TRIGONOMETRY PROOF

 PROVE SIN2A+SIN2B-SINC=4COSACOSBSINC

6 years ago

Answers : (2)

Dear bayana,
he double angle formula:
sin 2Θ = 2 sin Θ cos Θ

sin 2A + sin 2B - sin 2C
... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C 

Since A + B + C = π ;
A is a supplement angle of ( B + C )
B is a supplement angle of ( A + C )
C is a supplement angle of ( A + B )
TAKE NOTE that the sine of supplementary angles are equal !!!

sin 2A + sin 2B - sin 2C
... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C 
... = 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C

From the Sum of Angle Identity:
sin ( α + ß ) = sin α cos ß + cos α sin ß

sin 2A + sin 2B - sin 2C
... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C 
... = 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C
... = 2 ( sin B cos C + cos B sin C ) cos A 
..... ..... + 2 ( sin A cos C + cos A sin C ) cos B
..... ..... – 2 ( sin A cos B + cos A sin B ) cos C
... = 2 cos A sin B cos C + 2 cos A cos B sin C
..... ..... + 2 sin A cos B cos C + 2 cos A cos B sin C
..... ..... – 2 sin A cos B cos C – 2 cos A sin B cos C
... = 2 cos A cos B sin C + 2 cos A cos B sin C
... = 4 cos A cos B sin C

Answer 2

Answer:

Answer is below

Step-by-step explanation: