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Question: Prove that \(\sin 3x+\sin 2x-\sin x=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}\)....

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}.

Explanation

Solution

Hint: Use the trigonometric formulas of (sin⁡A−sin⁡B)\left( \sin A-\sin B \right), value of sin⁡2x\sin 2xto solve the expression. Take LHS of the expression and prove that the simplified expression of LHS is equal to RHS.

Complete step-by-step solution -
Given the expression, 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 first take the LHS.
LHS=sin⁡3x+sin⁡2x−sin⁡xLHS =\sin 3x +\sin 2x-\sin x.
We know the formula, sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)
Let us put, A = 2x and B = x.
∴sin⁡3x+(sin⁡2x−sin⁡x)=sin⁡3x+2cos⁡(2x+x2)sin⁡(2x−x2)\therefore \sin 3x+\left( \sin 2x-\sin x \right)=\sin 3x+2\cos \left( \dfrac{2x+x}{2} \right)\sin \left( \dfrac{2x-x}{2} \right)
=sin⁡3x+2cos⁡(3x2)sin⁡(x2)=\sin 3x+2\cos \left( \dfrac{3x}{2} \right)\sin \left( \dfrac{x}{2} \right)
We know that, sin⁡2x=2sin⁡x.cos⁡x\sin 2x=2\sin x.\cos x.
Similarly, sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}.
Now let us replace x with 3x.
∴sin⁡3x=2sin⁡(3x2)cos⁡(3x2) ∴LHS=2sin⁡(3x2)cos⁡(3x2)+2cos⁡(3x2)sin⁡(x2) \begin{aligned} & \therefore \sin 3x=2\sin \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{3x}{2} \right) \\\ & \therefore LHS=2\sin \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{3x}{2} \right)+2\cos \left( \dfrac{3x}{2} \right)\sin \left( \dfrac{x}{2} \right) \\\ \end{aligned}
=2cos⁡(3x2)[sin⁡(3x2)+sin⁡(x2)]=2\cos \left( \dfrac{3x}{2} \right)\left[ \sin \left( \dfrac{3x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right]
We know that, sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right).
Put, A=3x2A=\dfrac{3x}{2}and B=x2B=\dfrac{x}{2}.
∴LHS=2cos⁡(3x2)[2sin⁡(3x2+x22)cos⁡(3x2−x22)]\therefore LHS=2\cos \left( \dfrac{3x}{2} \right)\left[ 2\sin \left( \dfrac{\dfrac{3x}{2}+\dfrac{x}{2}}{2} \right)\cos \left( \dfrac{\dfrac{3x}{2}-\dfrac{x}{2}}{2} \right) \right]
=2cos⁡(3x2)[2sin⁡(4x4)cos⁡(2x4)] =2cos⁡(3x2)[2sin⁡xcos⁡(x2)] =4cos⁡(3x2)sin⁡xcos⁡x2 =4sin⁡xcos⁡(x2)cos⁡(3x2) \begin{aligned} & =2\cos \left( \dfrac{3x}{2} \right)\left[ 2\sin \left( \dfrac{4x}{4} \right)\cos \left( \dfrac{2x}{4} \right) \right] \\\ & =2\cos \left( \dfrac{3x}{2} \right)\left[ 2\sin x\cos \left( \dfrac{x}{2} \right) \right] \\\ & =4\cos \left( \dfrac{3x}{2} \right)\sin x\cos \dfrac{x}{2} \\\ & =4\sin x\cos \left( \dfrac{x}{2} \right)\cos \left( \dfrac{3x}{2} \right) \\\ \end{aligned}
Hence, we proved that LHS = RHS.
Here from the question, RHS=4cos⁡(x2)sin⁡xcos⁡(3x2)RHS=4\cos \left( \dfrac{x}{2} \right)\sin x\cos \left( \dfrac{3x}{2} \right).
Thus, we have proved.

Note: Remember the trigonometric formulae and identities to solve problems like these. Take (sin⁡2x−sin⁡x)\left( \sin 2x-\sin x \right)first because you will get a simplified expression and will be able to solve the expression more easily. We have used formulas like (sin⁡A−sin⁡B)\left( \sin A-\sin B \right), sin⁡2x,(sin⁡A+sin⁡B)\sin 2x,\left( \sin A+\sin B \right) to solve the expression. Therefore, remember the formulae.