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

Prove the following:
sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2sin⁡x\dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x

Explanation

Solution

We can take the LHS of the given equation. Then we can simplify its numerator using the trigonometric identity sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right). We can simplify the denominator using the identity cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = {\cos ^2}A - {\sin ^2}A. On doing further calculations, we will obtain the RHS of the equation. We can say the given equation is true when L.H.S=R.H.S

Complete step-by-step answer:
We need to prove that sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2sin⁡x\dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x
Let us look at the LHS,
LHS=sin⁡x−sin⁡3xsin⁡2x−cos⁡2xLHS = \dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} … (1)
We can consider the numerator of the LHS
We know that sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
We can substitute the values,
⇒sin⁡(x)−sin⁡(3x)=2cos⁡(x+3x2)sin⁡(x−3x2)\Rightarrow \sin \left( x \right) - \sin \left( {3x} \right) = 2\cos \left( {\dfrac{{x + 3x}}{2}} \right)\sin \left( {\dfrac{{x - 3x}}{2}} \right)
On simplification, we get,
⇒sin⁡(x)−sin⁡(3x)=2cos⁡(2x)sin⁡(−x)\Rightarrow \sin \left( x \right) - \sin \left( {3x} \right) = 2\cos \left( {2x} \right)\sin \left( { - x} \right)
We know thatsin⁡(−x)=−sin⁡(x)\sin \left( { - x} \right) = - \sin \left( x \right). So, we get,
⇒sin⁡(x)−sin⁡(3x)=−2cos⁡(2x)sin⁡(x)\Rightarrow \sin \left( x \right) - \sin \left( {3x} \right) = - 2\cos \left( {2x} \right)\sin \left( x \right)… (2)
We can consider the denominator of the LHS
We know that cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = {\cos ^2}A - {\sin ^2}A
We can multiply both sides with -1.
−cos⁡2A=sin⁡2A−cos⁡2A- \cos 2A = {\sin ^2}A - {\cos ^2}A
We can substitute the values,
sin⁡2x−cos⁡2x=−cos⁡2x{\sin ^2}x - {\cos ^2}x = - \cos 2x… (3)
We can substitute (3) and (2) in (1)
⇒LHS=−2cos⁡(2x)sin⁡(x)−cos⁡2x\Rightarrow LHS = \dfrac{{ - 2\cos \left( {2x} \right)\sin \left( x \right)}}{{ - \cos 2x}}
On further simplification, we get,
⇒LHS=2sin⁡x\Rightarrow LHS = 2\sin x
RHS is also equal to 2sin⁡x2\sin x. So, we can write,
L.H.S=R.H.S
Hence the equation is proved.

Note: We must be familiar with the following trigonometric identities used in this problem.
1.cos⁡(A)+cos⁡(B)=2cos⁡(A+B2)cos⁡(A−B2)\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
2.cos⁡(A)−cos⁡(B)=−2sin⁡(A+B2)sin⁡(A−B2)\cos \left( A \right) - \cos \left( B \right) = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
3.sin⁡(A)+sin⁡(B)=2sin⁡(A+B2)cos⁡(A−B2)\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
4.sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
5.cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = {\cos ^2}A - {\sin ^2}A
6.sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A
7.sin⁡(−x)=−sin⁡(x)\sin \left( { - x} \right) = - \sin \left( x \right)
8.cos⁡(−x)=cos⁡(x)\cos \left( { - x} \right) = \cos \left( x \right)
We must know the values of trigonometric functions at common angles. Adding π\pi or multiples of π\pi with the angle retains the ratio and adding π2\dfrac{\pi }{2}or odd multiples of π2\dfrac{\pi }{2}will change the ratio. While converting the angles we must take care of the sign of the ratio in its respective quadrant. In the 1st quadrant all the trigonometric ratios are positive. In the 2nd quadrant only sine and sec are positive. In the third quadrant, only tan and cot are positive and in the fourth quadrant, only cos and sec are positive. The angle measured in the counter clockwise direction is taken as positive and angle measured in the clockwise direction is taken as negative.