If A, B, C are angles of a triangle, then sin | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) $\sin 2A + \sin 2B\, - \sin 2C$ 

$= 2\sin A\cos A + 2\cos (B + C)\sin (B - C)$

$\{ \because A + B + C = \pi ,\,	herefore \,B

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $\sin 2A + \sin 2B\, - \sin 2C$ 

$= 2\sin A\cos A + 2\cos (B + C)\sin (B - C)$

$\{ \because A + B + C = \pi ,\,	herefore \,B + C = \pi  - A,\cos (B + C) = \cos (\pi  - A),$ $\cos (B + C) = - \cos A,\,\sin (B + C) = \sin A\} $

$ = 2\cos A\,\,[\sin A - \sin (B - C)]$

$ = 2\cos A\,[\sin (B + C) - \sin (B - C)]$ 

$ = 2\cos A.2\cos B.\sin C$

$ = 4\cos A.\,\cos B.\,\sin C$.

Trick : First put $A = B = C = {60^o}$ for, these values. 

Options $(a)$ and $ (d)$ satisfies the condition, Now put $A = B = 45^\circ $ and $c = {90^o}$. 

Then only $(d)$ satisfies. Hence $(d)$ is the answer.

If $A, B, C$ are angles of a triangle, then $\sin 2A + \sin 2B - \sin 2C$ is equal to

Similar Questions

Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$

If $\tan A = \frac{{1 - \cos B}}{{\sin B}},$ find $\tan 2A$ in terms of $\tan B$ and show that

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

If $\frac{{\cos x}}{a} = \frac{{\cos (x + \theta )}}{b} = \frac{{\cos (x + 2\theta )}}{c} = \frac{{\cos (x + 3\theta )}}{d} \, ,$ then $\left( {\frac{{a + c}}{{b + d}}} \right)$ is equal to :-

The expression $\frac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$ is equal to