Question

In triangle ABC, the value of $\sin 2A + \sin 2B + \sin 2C$ is equal to
A. $4\sin A\sin B\sin C$
B. $4\cos A\cos B\cos C$
C. $2\cos A\cos B\cos C$
D. $2\sin A\sin B\sin C$

Answer 1

Hint : In this type of question the key observation is to use the angle sum property of a triangle.Sum of all angles in a triangle is 180 degree. Also the trigonometry identities like$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ are used to solve this question.

Complete step-by-step answer :
In a $\Delta ABC$, we know that
$\angle A + \angle B + \angle C = {180^ \circ }$
Subtracting $\angle C$ on both sides,
$\angle A + \angle B = {180^ \circ } - \angle C$
Taking sin both sides,
$\sin \left( {A + B} \right) = \sin \left( {{{180}^ \circ } - C} \right)$
$\because \sin \left( {{{180}^ \circ } - C{\text{ }}} \right){\text{ lies in 2nd quadrant }}$
$\therefore \sin \left( {A + B} \right) = \sin C$
The equation given is,
Let $I = \sin 2A + \sin 2B + \sin 2C$
$\because \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\Rightarrow I = 2\sin \left( {\dfrac{{2\left( {A + B} \right)}}{2}} \right)\cos \left( {\dfrac{{2\left( {A - B} \right)}}{2}} \right) + \sin 2C$
On simplifying further,
$\Rightarrow I = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + \sin 2C$
$\because \sin \left( {A + B} \right) = \sin C$
$\Rightarrow I = 2\sin \left( C \right)\cos \left( {A - B} \right) + \sin 2C$
Using $\sin 2x = 2\sin x\cos x$
$\Rightarrow I = 2\sin \left( C \right)\cos \left( {A - B} \right) + 2\sin C\cos C$
On taking 2sinC common,
$I = 2\sin \left( C \right)\left( {\cos \left( {A - B} \right) + \cos C} \right)$
Again in a $\Delta ABC$,
$\angle A + \angle B + \angle C = {180^ \circ }$
Subtracting $\angle C$ on both sides,
$\angle A + \angle B = {180^ \circ } - \angle C$
Taking cos both sides,
$\cos \left( {A + B} \right) = \cos \left( {{{180}^ \circ } - C} \right)$
$\because \cos \left( {{{180}^ \circ } - C{\text{ }}} \right){\text{ lies in 2nd quadrant }}$
$\Rightarrow \cos \left( {A + B} \right) = - \cos C$
Or,
$\cos C = - \cos \left( {A + B} \right)$
On putting $\cos C = - \cos \left( {A + B} \right)$ in I,
$\Rightarrow I = 2\sin \left( C \right)\left( {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right)$
$\because \cos A + \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\therefore {\text{ On applying,}}$
$\Rightarrow I = 2\sin \left( C \right)\left( { - 2\sin \dfrac{{A - B + \left( {A + B} \right)}}{2}\sin \dfrac{{A - B - \left( {A + B} \right)}}{2}} \right)$
On simplifying,
$\Rightarrow I = 2\sin \left( C \right)\left( { - 2\sin A\sin \left( { - B} \right)} \right)$
$\because \sin \left( { - x} \right) = - \sin x$
$\Rightarrow I = 2\sin \left( C \right)\left( {2\sin A\sin B} \right)$
On simplifying further,
$I = 4\sin A\sin B\sin C$
So, the correct answer is “Option A”.

Note : Remember that in first quadrant all trigonometry functions are positive, in second quadrant only $\sin \theta$ is positive, in third quadrant only $\tan \theta$ is positive and in fourth quadrant only $\cos \theta$ is positive. Calculations should be done carefully to avoid any mistake. After the final answer is found out it can be checked that whether it satisfies the original equation given in the question by simply substituting its value in the equation and if it does not satisfy the equation then the solution must be rechecked. The equation should be solved in accordance with the identities which would result in the correct solution.