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3.Trigonometrical Ratios, Functions and Identities
medium
જો $A + B + C = \frac{{3\pi }}{2},$ તો $\cos 2A + \cos 2B + \cos 2C = $
A
$1 - 4\cos A\,\cos B\,\cos C$
B
$4\sin A\,\,\sin B\,\,\sin C$
C
$1 + 2\cos A\,\cos B\,\cos C$
D
$1 - 4\sin A\,\,\sin B\,\,\sin C$
Solution
(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 $A = B = C = \frac{\pi }{2}$.
Standard 11
Mathematics
