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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

Hint:If you look carefully, the numerator of the L.H.S expression is in the form of sin⁡C−sin⁡D\sin C-\sin D and the denominator of the L.H.S expression is in the form of cos⁡2θ−sin⁡2θ{{\cos }^{2}}\theta -{{\sin }^{2}}\theta . So, apply the identity ofsin⁡C−sin⁡D\sin C-\sin D and cos⁡2θ−sin⁡2θ{{\cos }^{2}}\theta -{{\sin }^{2}}\theta in the numerator and denominator of L.H.S respectively and then solve.

Complete step-by-step answer:
We have to prove the following expression:
sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2sin⁡x\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}=2\sin x
Solving L.H.S of the expression,
sin⁡x−sin⁡3xsin⁡2x−cos⁡2x\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}
The numerator of the above expression is in the form of sin⁡C−sin⁡D\sin C-\sin D and 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)
Applying this formula in sin⁡x−sin⁡3x\sin x-\sin 3x we get,
sin⁡x−sin⁡3x=2cos⁡2xsin⁡(−x)\sin x-\sin 3x=2\cos 2x\sin (-x)
Now, the denominator of the L.H.S expression is in the form of cos⁡2θ−sin⁡2θ{{\cos }^{2}}\theta -{{\sin }^{2}}\theta and we know that:
cos⁡2θ−sin⁡2θ=cos⁡2θ{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta
Applying this formula in sin⁡2x−cos⁡2x{{\sin }^{2}}x-{{\cos }^{2}}x we get,
sin⁡2x−cos⁡2x=−cos⁡2x{{\sin }^{2}}x-{{\cos }^{2}}x=-\cos 2x
Substituting the value of sin⁡x−sin⁡3x\sin x-\sin 3x and sin⁡2x−cos⁡2x{{\sin }^{2}}x-{{\cos }^{2}}x in the L.H.S expression of the given question we get,
sin⁡x−sin⁡3xsin⁡2x−cos⁡2x\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}
=2cos⁡2xsin⁡(−x)−cos⁡2x=\dfrac{2\cos 2x\sin (-x)}{-\cos 2x}
From the trigonometric properties we know that:
sin⁡(−x)=−sin⁡x\sin (-x)=-\sin x.
Now, we are going to substitute this value in the above expression and cos⁡2x\cos 2x will be cancelled out from the numerator and the denominator of the above expression.
−2sin⁡x−1 =2sin⁡x \begin{aligned} & \dfrac{-2\sin x}{-1} \\\ & =2\sin x \\\ \end{aligned}
The answer we get from solving the L.H.S of the given expression is 2sin⁡x2\sin x which is equal to R.H.S.
Hence, we have proved that L.H.S = R.H.S of the given expression.

Note: You can do the above proof by cross – multiplying the given equation:
sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2sin⁡x\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}=2\sin x
sin⁡x−sin⁡3x=2sin⁡x(sin⁡2x−cos⁡2x)\sin x-\sin 3x=2\sin x\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)
Multiplying -1 on both the sides we get,
sin⁡3x−sin⁡x=2sin⁡x(cos⁡2x−sin⁡2x)\sin 3x-\sin x=2\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)…………. Eq. (1)
Solving R.H.S of the above equation:
2sin⁡x(cos⁡2x−sin⁡2x)2\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)
We know that cos⁡2x−sin⁡2x=cos⁡2x{{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x so using this relation in the above expression.
2sin⁡xcos⁡2x2\sin x\cos 2x
Now, solving R.H.S of the eq. (1) we get,
sin⁡3x−sin⁡x\sin 3x-\sin x
Using the formula of sin⁡C−sin⁡D\sin C-\sin D in the above equation we get,
sin⁡3x−sin⁡x=2cos⁡2xsin⁡(x)\sin 3x-\sin x=2\cos 2x\sin (x)
The above simplification of L.H.S of eq. (1) has given the value 2cos⁡2xsin⁡x2\cos 2x\sin x and R.H.S we have solved above has given the value 2sin⁡xcos⁡2x2\sin x\cos 2x.
As you can see L.H.S = R.H.S. so we have proved.