Question

Prove the following:
\[\sin 3x+\sin 2x-\sin x=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}\]

Answer 1

Hint: In this question, consider the LHS and rearrange the expression as (sin 3x – sin x) + sin 2x. Now use \[\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)\] and \[\sin 2\theta =2\sin \theta \cos \theta \] and then take the common terms out and use \[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\] to prove the desired result.
Complete step-by-step answer:
Here, we have to prove that
\[\sin 3x+\sin 2x-\sin x=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}\]
Let us consider the LHS of the expression given in the question.
\[E=\sin 3x+\sin 2x-\sin x\]
By rearranging the terms of the above expression, we get,
\[E=\left( \sin 3x-\sin x \right)+\sin \left( 2x \right)\]
We know that \[\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)\]. By using this in the above expression, we get,
\[E=2\cos \left( \dfrac{3x+x}{2} \right)\sin \left( \dfrac{3x-x}{2} \right)+\sin \left( 2x \right)\]
\[E=2\cos \left( \dfrac{4x}{2} \right)\sin \left( \dfrac{2x}{2} \right)+\sin \left( 2x \right)\]
\[E=2\cos \left( 2x \right)\sin \left( x \right)+\sin \left( 2x \right)\]
Now, we know that \[\sin 2\theta =2\sin \theta \cos \theta \]. By using this in the above expression, we get,
\[E=2\cos \left( 2x \right)\sin \left( x \right)+2\sin \left( x \right)\cos \left( x \right)\]
By taking out 2 sin x common from the above expression, we get,
\[E=2\sin x\left( \cos 2x+\cos x \right)\]
We know that,
\[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]
By using this in the above expression, we get,
\[E=2\sin x\left[ 2\cos \left( \dfrac{2x+x}{2} \right)\cos \left( \dfrac{2x-x}{2} \right) \right]\]
\[E=2\sin x\left[ 2\cos \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{x}{2} \right) \right]\]
\[E=4\sin x\cos \dfrac{3x}{2}\cos \dfrac{x}{2}\]
We can also write the above expression as,
\[E=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}\]
E = RHS
So, we get, LHS = RHS.
Hence, we have proved that
\[\sin 3x+\sin 2x-\sin x=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}\]

Note: In this question, first of all, students need to select which terms to the club first. Suppose, in this question, if we would have clubbed sin 3x + sin 2x first and used \[\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\], we would not be able to prove the desired result because \[\dfrac{3x+2x}{2}=\dfrac{5x}{2}\] which is not there in the RHS. So, students should first properly look at RHS and then only proceed to solve the LHS to prove the desired result.