In triangle ABC, the value of sin 2A + sin 2B | vedclass

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Question

(a) We know that $A + B + C = 180^\circ $ (in $\Delta ABC$)

Now, $\sin 2A + \sin 2B + \sin 2C$

$ = 2\sin (A + B)\cos (A - B) + 2\sin C\cos C$

Vedclass · Accepted Answer

$4\sin A.\,\sin B.\,\sin C$

Vedclass · Answer

(a) We know that $A + B + C = 180^\circ $ (in $\Delta ABC$)

Now, $\sin 2A + \sin 2B + \sin 2C$

$ = 2\sin (A + B)\cos (A - B) + 2\sin C\cos C$

$ = 2\sin (\pi - C)\cos (A - B) + 2\sin C\cos (\pi - \overline {A + B} )$

$ = 2\sin C\cos (A - B) - 2\sin C\cos (A + B)$ 

$ = 2\sin C\{ \cos (A - B) - \cos (A + B)\} $

$ = 2\sin C\{ 2\sin A\sin B\} = 4\sin A\sin B\sin C$.

In triangle $ABC$, the value of $\sin 2A + \sin 2B + \sin 2C$ is equal to

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