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If A, B, and C are the interior angles of triangle ABC, prove that

4 sinA sinB sinC = sin2A + sin2B + sin2C

User Clh
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1 Answer

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NOTES:

1)⇒ A + B + C = 180°

A + B + C = π

A + B = π - C

2)⇒⇒ sin (A + B) = sin (π - C)

= (sin π)(cos C) - (sin C)(cos π)

= (0)(cos C) - (sin C)(-1)

= 0 - (-sin C)

= sin C

3)⇒⇒⇒cos (A + B) = cos (π - c)

= (cos π)(cos C) + (sin π)(sin C)

= (-1)(cos C) + (0)(sin C)

= - cos C

4)⇒⇒⇒⇒ sin 2A + sin 2B = 2 sin (A + B) cos (A - B)

PROOF (from left side):

sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

2 sin (A + B) cos (A - B) + sin 2C refer to NOTE 4

2 sin (A + B) cos (A - B) + 2 sin C cos C double angle formula

2 sin C cos (A - B) + 2 sin C cos C refer to NOTE 2

2 sin C [cos (A - B) + cos C] factored out 2 sin C

2 sin C [cos (A - B) - (cos(A + B)] refer to NOTE 3

2 sin C [2 sin A sin B] sum/difference formula

4 sin A sin B sin C multiplied 2 sin C by 2 sin A sin B


Proof completed: 4 sin A sin B sin C = 4 sin A sin B sin C



User Nobik
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