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

Prove the following:
sin⁡3x+sin⁡2x−sin⁡x=4sin⁡xcos⁡x2cos⁡3x2\sin 3x+\sin 2x-\sin x=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}

Explanation

Solution

Hint: In this question, consider the LHS and rearrange the expression as (sin 3x – sin x) + sin 2x. Now use sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) and sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta and then take the common terms out and use cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\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=4sin⁡xcos⁡x2cos⁡3x2\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⁡xE=\sin 3x+\sin 2x-\sin x
By rearranging the terms of the above expression, we get,
E=(sin⁡3x−sin⁡x)+sin⁡(2x)E=\left( \sin 3x-\sin x \right)+\sin \left( 2x \right)
We know that sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\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=2cos⁡(3x+x2)sin⁡(3x−x2)+sin⁡(2x)E=2\cos \left( \dfrac{3x+x}{2} \right)\sin \left( \dfrac{3x-x}{2} \right)+\sin \left( 2x \right)
E=2cos⁡(4x2)sin⁡(2x2)+sin⁡(2x)E=2\cos \left( \dfrac{4x}{2} \right)\sin \left( \dfrac{2x}{2} \right)+\sin \left( 2x \right)
E=2cos⁡(2x)sin⁡(x)+sin⁡(2x)E=2\cos \left( 2x \right)\sin \left( x \right)+\sin \left( 2x \right)
Now, we know that sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta . By using this in the above expression, we get,
E=2cos⁡(2x)sin⁡(x)+2sin⁡(x)cos⁡(x)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=2sin⁡x(cos⁡2x+cos⁡x)E=2\sin x\left( \cos 2x+\cos x \right)
We know that,
cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\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=2sin⁡x[2cos⁡(2x+x2)cos⁡(2x−x2)]E=2\sin x\left[ 2\cos \left( \dfrac{2x+x}{2} \right)\cos \left( \dfrac{2x-x}{2} \right) \right]
E=2sin⁡x[2cos⁡(3x2)cos⁡(x2)]E=2\sin x\left[ 2\cos \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{x}{2} \right) \right]
E=4sin⁡xcos⁡3x2cos⁡x2E=4\sin x\cos \dfrac{3x}{2}\cos \dfrac{x}{2}
We can also write the above expression as,
E=4sin⁡xcos⁡x2cos⁡3x2E=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=4sin⁡xcos⁡x2cos⁡3x2\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=2sin⁡(C+D2)cos⁡(C−D2)\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 3x+2x2=5x2\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.