Question

If we have the sum of angles of a triangle as $A+B+C=0$, then $\sin A+\sin B+\sin C=$
(a) $2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$
(b) $-2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$
(c) $4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$
(d) $-4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$

Answer 1

Hint: We will rearrange the given equation as $A=-\left( B+C \right)$. We will use the identity for trigonometric functions for addition of angles. This identity is as follows,
$\sin \left( x+y \right)=\sin x\cos y + \cos x\sin y$.
We will use the double angle identity for the cosine function which is given as $\cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2}$. We will also use the double angle identity for the sine function which is $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$. Using these identities we will be able to simplify the given expression to obtain the desired answer.

Complete step by step solution:
We know that $A+B+C=0$, so we can write it as $A=-\left( B+C \right)$. Now, we have to simplify the following expression: $\sin A+\sin B + \sin C$. Since $A=-\left( B+C \right)$, we can write $\sin A=\sin \left( -\left( B+C \right) \right)$. We are aware of the trigonometric functions for negative angles. We will use $\sin \left( -x \right)=-\sin x$. So we have $\sin A=-\sin \left( B+C \right)$. We know that $\sin \left( x+y \right)=\sin x\cos y + \cos x\sin y$, which is the identity for trigonometric function for addition of angles. Using this identity, we get
$\begin{align}
  & \sin A=-\sin \left( B+C \right) \\
 & =-\left[ \sin B\cos C+\cos B\sin C \right] \\
 & =-\sin B\cos C-\cos B\sin C
\end{align}$
We will substitute this expression of $\sin A$ in the given expression in the following manner,
$\sin A+\sin B +\sin C=-\sin B\cos C-\cos B\sin C+\sin B +\sin C$
We can take $\sin B$ and $\sin C$ as common factors in the following manner,
$\sin A+\sin B + \sin C=\sin B\left( 1-\cos C \right)+\sin C\left( 1-\cos B \right)$.
We know that $\cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2}$. So, we can write $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$. Substituting this in the above expression, we get
$\sin A+\sin B + \sin C=\sin B\left( 2{{\sin }^{2}}\dfrac{C}{2} \right)+\sin C\left( 2{{\sin }^{2}}\dfrac{B}{2} \right)$.
Now we will use the following double angle identity for the sine function, $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$.
So, the above expression will become
$\sin A+\sin B + \sin C=2\sin \dfrac{B}{2}\cos \dfrac{B}{2}\left( 2{{\sin }^{2}}\dfrac{C}{2} \right)+2\sin \dfrac{C}{2}\cos \dfrac{C}{2}\left( 2{{\sin }^{2}}\dfrac{B}{2} \right)$
Rearranging the RHS of the above expression, we get
$\sin A+\sin B+ \sin C=4\sin \dfrac{B}{2}\sin \dfrac{C}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2}+4\sin \dfrac{C}{2}\sin \dfrac{B}{2}\cos \dfrac{C}{2}\sin \dfrac{B}{2}$
Taking $4\sin \dfrac{B}{2}\sin \dfrac{C}{2}$ as a common factor, we get
$\sin A+\sin B+\sin C=4\sin \dfrac{B}{2}\sin \dfrac{C}{2}\left[ \cos \dfrac{B}{2}\sin \dfrac{C}{2}+\cos \dfrac{C}{2}\sin \dfrac{B}{2} \right]$
Using the identity for trigonometric function for addition of angles, we get
$\left[ \cos \dfrac{B}{2}\sin \dfrac{C}{2}+\cos \dfrac{C}{2}\sin \dfrac{B}{2} \right]=\sin \left( \dfrac{B}{2}+\dfrac{C}{2} \right)=\sin \left( \dfrac{B+C}{2} \right)$
Substituting this in the above expression, we get
$\sin A+\sin B+\sin C=4\sin \dfrac{B}{2}\sin \dfrac{C}{2}\sin \left( \dfrac{B+C}{2} \right)$
But we know that $A=-\left( B+C \right)$, so $B+C=-A$. Hence, we get
$\sin A+\sin B+ \sin C=4\sin \dfrac{B}{2}\sin \dfrac{C}{2}\sin \left( \dfrac{-A}{2} \right)$
Again, using the identity $\sin \left( -x \right)=-\sin x$, we get
$\sin A+\sin B+\sin C=-4\sin \dfrac{B}{2}\sin \dfrac{C}{2}\sin \dfrac{A}{2}$
Hence, the correct option is (d).

Note: It is essential that we are familiar with the trigonometric identities for double angles, the sum of angles, etc. By looking at the given options, we should get the hint that we will need to use either double angle or half-angle formulae. It is important to keep a check of the signs for every expression since there is a possibility that we might misplace the signs which will lead to incorrect answers.