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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...

If we have the sum of angles of a triangle as A+B+C=0A+B+C=0, then sin⁡A+sin⁡B+sin⁡C=\sin A+\sin B+\sin C=
(a) 2sin⁡A2sin⁡B2sin⁡C22\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}
(b) −2sin⁡A2sin⁡B2sin⁡C2-2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}
(c) 4sin⁡A2sin⁡B2sin⁡C24\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}
(d) −4sin⁡A2sin⁡B2sin⁡C2-4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}

Explanation

Solution

We will rearrange the given equation as A=−(B+C)A=-\left( B+C \right). We will use the identity for trigonometric functions for addition of angles. This identity is as follows,
sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\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⁡θ=1−2sin⁡2θ2\cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2}. We will also use the double angle identity for the sine function which is sin⁡θ=2sin⁡θ2cos⁡θ2\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=0A+B+C=0, so we can write it as A=−(B+C)A=-\left( B+C \right). Now, we have to simplify the following expression: sin⁡A+sin⁡B+sin⁡C\sin A+\sin B + \sin C. Since A=−(B+C)A=-\left( B+C \right), we can write sin⁡A=sin⁡(−(B+C))\sin A=\sin \left( -\left( B+C \right) \right). We are aware of the trigonometric functions for negative angles. We will use sin⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x. So we have sin⁡A=−sin⁡(B+C)\sin A=-\sin \left( B+C \right). We know that sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\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
sin⁡A=−sin⁡(B+C) =−[sin⁡Bcos⁡C+cos⁡Bsin⁡C] =−sin⁡Bcos⁡C−cos⁡Bsin⁡C\begin{aligned} & \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{aligned}
We will substitute this expression of sin⁡A\sin A in the given expression in the following manner,
sin⁡A+sin⁡B+sin⁡C=−sin⁡Bcos⁡C−cos⁡Bsin⁡C+sin⁡B+sin⁡C\sin A+\sin B +\sin C=-\sin B\cos C-\cos B\sin C+\sin B +\sin C
We can take sin⁡B\sin B and sin⁡C\sin C as common factors in the following manner,
sin⁡A+sin⁡B+sin⁡C=sin⁡B(1−cos⁡C)+sin⁡C(1−cos⁡B)\sin A+\sin B + \sin C=\sin B\left( 1-\cos C \right)+\sin C\left( 1-\cos B \right).
We know that cos⁡θ=1−2sin⁡2θ2\cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2}. So, we can write 1−cos⁡θ=2sin⁡2θ21-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}. Substituting this in the above expression, we get
sin⁡A+sin⁡B+sin⁡C=sin⁡B(2sin⁡2C2)+sin⁡C(2sin⁡2B2)\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⁡θ=2sin⁡θ2cos⁡θ2\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}.
So, the above expression will become
sin⁡A+sin⁡B+sin⁡C=2sin⁡B2cos⁡B2(2sin⁡2C2)+2sin⁡C2cos⁡C2(2sin⁡2B2)\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=4sin⁡B2sin⁡C2cos⁡B2sin⁡C2+4sin⁡C2sin⁡B2cos⁡C2sin⁡B2\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 4sin⁡B2sin⁡C24\sin \dfrac{B}{2}\sin \dfrac{C}{2} as a common factor, we get
sin⁡A+sin⁡B+sin⁡C=4sin⁡B2sin⁡C2[cos⁡B2sin⁡C2+cos⁡C2sin⁡B2]\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
[cos⁡B2sin⁡C2+cos⁡C2sin⁡B2]=sin⁡(B2+C2)=sin⁡(B+C2)\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=4sin⁡B2sin⁡C2sin⁡(B+C2)\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=−(B+C)A=-\left( B+C \right), so B+C=−AB+C=-A. Hence, we get
sin⁡A+sin⁡B+sin⁡C=4sin⁡B2sin⁡C2sin⁡(−A2)\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⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x, we get
sin⁡A+sin⁡B+sin⁡C=−4sin⁡B2sin⁡C2sin⁡A2\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.