Question

If A+B+C=90degrees then sin2A+sin2B+sin2C=?

Options 1)cosAcosBcosC

2)2cosAcosBcosC

3)3cosAcosBcosC

4)4cosAcosBcosC

Answer is option 4

Answer 1

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• TRIGONOMETRY PROOF

The question is SIN2A+SIN2B-SINC

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



Now taking sin2A+sin2B +sin2C
= If A+B+C=180 degree, then A+B=180-C hence sin(A+B) = sinC, cos(A+B)=-cosC

sin2A + sin 2B +sin2C

= 2 sin(A+B)cos(A-B) + 2sinC cosC

=2sinC cos(A-B)+2sinC cosC

=2sinC (cos(A-b) + cos C)

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

= 2sinC . 2sin A sin B

=4 sinA sin B sin C