Answer
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Hint: In this question, apply (C, D) formula on the first two terms and apply double angle formula on the third term. Also apply suitable transformation formula upon the terms inside the bracket.
Complete step-by-step answer:
We write here the steps, which we follow in general-
1. Given $A+B+C=\pi \Rightarrow A+B=\pi -C$
2. Apply (C, D) formula on the first two terms and apply double angle formula on the third term.
3. Replace $A+B$ by $\pi -C$ inside the bracket.
4. Apply suitable transformation formula upon the terms inside the bracket.
Let us consider the L. H. S.
L. H. S. $=\sin 2A+\sin 2B+\sin 2C$
L. H. S. $=\left( \sin 2A+\sin 2B \right)+\sin 2C$
Applying the sum to product formula $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ , we get
L. H. S. $=\left[ 2\sin (A+B)\cos (A-B) \right]+2\sin C\cos C.............(1)$
It is given that, $A+B+C=\pi \Rightarrow A+B=\pi -C$
So, the expression $\sin (A+B)=\sin (\pi -C)=\sin C$
Now put this expression in the equation (1), we get
L. H. S. $=\left[ 2\sin C\cos (A-B) \right]+2\sin C\cos C$
Rearranging the terms, we get
L. H. S. $=2\sin C\left[ \cos (A-B)+\cos C \right].......................(2)$
It is also given that, $C=\pi -(A+B)$
So, the expression $\cos C=\cos \left[ \pi -(A+B) \right]=-\cos (A+B)$
Now put this expression in the equation (2), we get
L. H. S. $=2\sin C\left[ \cos (A-B)-\cos (A+B) \right]$
Applying the formula$\cos (A+B)-\cos (A+B)=2\sin A\sin B$, we get
L. H. S. $=2\sin C\left[ 2\sin A\sin B \right]$
Rearranging the terms, we get
L. H. S. $=4\sin A\sin B\sin C$
L. H. S. = R. H. S.
Therefore, we have
$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
This is the desired result.
Note: This identity holds for all values of the angles which satisfy the given conditions among them and hence they are called conditional trigonometric identity. To establish such identity we require the properties of supplementary and complementary angles.
Complete step-by-step answer:
We write here the steps, which we follow in general-
1. Given $A+B+C=\pi \Rightarrow A+B=\pi -C$
2. Apply (C, D) formula on the first two terms and apply double angle formula on the third term.
3. Replace $A+B$ by $\pi -C$ inside the bracket.
4. Apply suitable transformation formula upon the terms inside the bracket.
Let us consider the L. H. S.
L. H. S. $=\sin 2A+\sin 2B+\sin 2C$
L. H. S. $=\left( \sin 2A+\sin 2B \right)+\sin 2C$
Applying the sum to product formula $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ , we get
L. H. S. $=\left[ 2\sin (A+B)\cos (A-B) \right]+2\sin C\cos C.............(1)$
It is given that, $A+B+C=\pi \Rightarrow A+B=\pi -C$
So, the expression $\sin (A+B)=\sin (\pi -C)=\sin C$
Now put this expression in the equation (1), we get
L. H. S. $=\left[ 2\sin C\cos (A-B) \right]+2\sin C\cos C$
Rearranging the terms, we get
L. H. S. $=2\sin C\left[ \cos (A-B)+\cos C \right].......................(2)$
It is also given that, $C=\pi -(A+B)$
So, the expression $\cos C=\cos \left[ \pi -(A+B) \right]=-\cos (A+B)$
Now put this expression in the equation (2), we get
L. H. S. $=2\sin C\left[ \cos (A-B)-\cos (A+B) \right]$
Applying the formula$\cos (A+B)-\cos (A+B)=2\sin A\sin B$, we get
L. H. S. $=2\sin C\left[ 2\sin A\sin B \right]$
Rearranging the terms, we get
L. H. S. $=4\sin A\sin B\sin C$
L. H. S. = R. H. S.
Therefore, we have
$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
This is the desired result.
Note: This identity holds for all values of the angles which satisfy the given conditions among them and hence they are called conditional trigonometric identity. To establish such identity we require the properties of supplementary and complementary angles.
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