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Question: Prove that: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \t...

Prove that: sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta .

Explanation

Solution

We take common trigonometric terms in numerator and denominator. We then find cos⁡2θ\cos 2\theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta using the trigonometric identities sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 and cos⁡2θ−sin⁡2θ=cos⁡2θ{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta . Then substitute in L.H.S of the given result and make necessary calculations to complete the required proof.

Complete step by step answer:
We have been given a trigonometric function which we need to prove that LHS = RHS = tan⁡θ\tan \theta .
We have been given that,
LHS = sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta } - (1)
From the above expression, let us take sin⁡θ\sin \theta common from the numerator and cos⁡θ\cos \theta as common from the denominator.
LHS = sin⁡θ(1−2sin⁡2θ)cos⁡θ(2cos⁡2θ−1)\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}- (2)
We know that,
cos⁡2θ−sin⁡2θ=cos⁡2θ{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta - (3)
We know that, sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.
From this above, sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta
cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta
Let us put sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta in (3). We get,
cos⁡2θ−1+cos⁡2θ=cos⁡2θ{{\cos }^{2}}\theta -1+{{\cos }^{2}}\theta =\cos 2\theta .
∴cos⁡2θ=2cos⁡2θ−1\therefore \cos 2\theta =2{{\cos }^{2}}\theta -1 - (4)
Now let us put cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta in (3). We get,
(1−sin⁡2θ)−sin⁡2θ=cos⁡2θ\left( 1-{{\sin }^{2}}\theta \right)-{{\sin }^{2}}\theta =\cos 2\theta
∴cos⁡2θ=1−2sin⁡2θ\therefore \cos 2\theta =1-2{{\sin }^{2}}\theta - (5)
Thus we got,

& \cos 2\theta =1-2{{\sin }^{2}}\theta \\\ & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\\ \end{aligned}$$ Let us put these expressions on (2). $$\therefore $$ LHS = $$\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}=\dfrac{\sin \theta .\cos 2\theta }{\cos \theta .\cos 2\theta }$$ Now let us cancel out $$\cos 2\theta $$ from the numerator and denominator. $$\therefore $$ LHS = $$\dfrac{\sin \theta }{\cos \theta }=\tan \theta $$. Thus we proved that, LHS = RHS = $$\tan \theta $$. Hence proved **Note:** From the numerator, don’t mistake it to be a formula for $$\sin 3x$$. Always try to simplify a trigonometric function. By simplifying it you will get an idea of how to solve the rest of the function. Remember the formula $${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $$, which is the key point for the solution. We should not make mistakes while solving this problem.