sin2A+sin2B-sin2C =4cosAcosBsinChow to solvethis question?
sin2A+sin2B-sin2C =4cosAcosBsinC
how to solvethis question?
Rayni , 7 Years ago
Grade 12
Trigonometry> sin2A+sin2B-sin2C =4cosAcosBsinC how to s...
1 Answers
Arun
A + B + C = π
sin 2A + sin 2B - sin 2C = 4 cos A cos B sin C
From the 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
sin 2A + sin 2B - sin 2C = 4 cos A cos B sin C
From the 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
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