Question

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

Answer 1

Hint: We can take the LHS of the given equation. Then we can simplify its numerator using the trigonometric identity \[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]. We can simplify the denominator using the identity $\cos 2A = {\cos ^2}A - {\sin ^2}A$. 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 \[\dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x\]
Let us look at the LHS,
$LHS = \dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}}$ … (1)
We can consider the numerator of the LHS
We know that \[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]
We can substitute the values,
 \[ \Rightarrow \sin \left( x \right) - \sin \left( {3x} \right) = 2\cos \left( {\dfrac{{x + 3x}}{2}} \right)\sin \left( {\dfrac{{x - 3x}}{2}} \right)\]
On simplification, we get,
\[ \Rightarrow \sin \left( x \right) - \sin \left( {3x} \right) = 2\cos \left( {2x} \right)\sin \left( { - x} \right)\]
We know that\[\sin \left( { - x} \right) = - \sin \left( x \right)\]. So, we get,
\[ \Rightarrow \sin \left( x \right) - \sin \left( {3x} \right) = - 2\cos \left( {2x} \right)\sin \left( x \right)\]… (2)
We can consider the denominator of the LHS
We know that $\cos 2A = {\cos ^2}A - {\sin ^2}A$
We can multiply both sides with -1.
$ - \cos 2A = {\sin ^2}A - {\cos ^2}A$
We can substitute the values,
 ${\sin ^2}x - {\cos ^2}x = - \cos 2x$… (3)
We can substitute (3) and (2) in (1)
$ \Rightarrow LHS = \dfrac{{ - 2\cos \left( {2x} \right)\sin \left( x \right)}}{{ - \cos 2x}}$
On further simplification, we get,
$ \Rightarrow LHS = 2\sin x$
RHS is also equal to \[2\sin x\]. 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.\[\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
2.\[\cos \left( A \right) - \cos \left( B \right) = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]
3.\[\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
4.\[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]
5.$\cos 2A = {\cos ^2}A - {\sin ^2}A$
6.$\sin 2A = 2\sin A\cos A$
7.\[\sin \left( { - x} \right) = - \sin \left( x \right)\]
8.\[\cos \left( { - x} \right) = \cos \left( x \right)\]
We must know the values of trigonometric functions at common angles. Adding $\pi $or multiples of $\pi $with the angle retains the ratio and adding $\dfrac{\pi }{2}$or odd multiples of $\dfrac{\pi }{2}$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.