If A + B + C = frac{{3pi }}{2}, then cos 2A + | vedclass

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Question

(d) $\cos 2A + \cos 2B + \cos 2C$ 

$ = 2\cos (A + B)\cos (A - B) + \cos 2C$ 

$ = 2\cos \left( {\frac{{3\pi }}{2} - C} \right)\cos (A -

Vedclass · Accepted Answer

$1 - 4\sin A\,\,\sin B\,\,\sin C$

Vedclass · Answer

(d) $\cos 2A + \cos 2B + \cos 2C$ 

$ = 2\cos (A + B)\cos (A - B) + \cos 2C$ 

$ = 2\cos \left( {\frac{{3\pi }}{2} - C} \right)\cos (A - B) + \cos 2C$

$ = - 2\sin C\cos (A - B) + 1 - 2{\sin ^2}C$

$ = 1 - 2\sin C\{ \cos (A - B) + \sin C\} $ 

$ = 1 - 2\sin C\left\{ {\cos (A - B) + \sin \left( {\frac{{3\pi }}{2} - (A + B)} \right)} \right\}$

$ = 1 - 2\sin C\{ \cos (A - B) - \cos (A + B)\} $

$ = 1 - 4\sin A\sin B\sin C$. 

Trick : Check by assuming $A = B = C = \frac{\pi }{2}$.

If $A + B + C = \frac{{3\pi }}{2},$ then $\cos 2A + \cos 2B + \cos 2C = $

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