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Question: Prove the equation: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }...

Prove the equation: 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

At first, take common sin⁡θ\sin \theta from the numerator and cos⁡θ\cos \theta from the denominator and after that, use the following identity cos⁡2θ{{\cos }^{2}}\theta equals to (1−2sin⁡2θ)\left( 1-2{{\sin }^{2}}\theta \right) or it can also be written as (2cos⁡2θ−1)\left( 2{{\cos }^{2}}\theta -1 \right) and hence, use the fact that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }.

Complete step by step solution:
In the given question, we have to prove that, the expression sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta } equals to or can be written as tan⁡θ\tan \theta .
As we are given the equation as,
sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta
So, we will first consider the left-hand side of the given equation and try to convert it into the expression of the right-hand side.
So, we have the left-hand side given as,
sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }
We will first take sin⁡θ\sin \theta common from the numerator and cos⁡θ\cos \theta common from the denominator so, we get,
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)}
Now, here we know an identity which is cos⁡2θ\cos 2\theta equals to 2cos⁡2θ−12{{\cos }^{2}}\theta -1 and also that cos⁡2θ\cos 2\theta equals to 1−2sin⁡2θ1-2{{\sin }^{2}}\theta .
So, applying the above identities, we get that the given expression further changes to,
sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=sin⁡θ×cos⁡2θcos⁡θ×cos⁡2θ\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\dfrac{\sin \theta \times \cos 2\theta }{\cos \theta \times \cos 2\theta }
It can also be written as,
sin⁡θcos⁡2θcos⁡θcos⁡2θ\dfrac{\sin \theta \cos 2\theta }{\cos \theta \cos 2\theta }
Now, by cancelling cos⁡2θ\cos 2\theta from both numerator and denominator, we get, the following expression as,
sin⁡θcos⁡2θcos⁡θcos⁡2θ=sin⁡θcos⁡θ\dfrac{\sin \theta \cos 2\theta }{\cos \theta \cos 2\theta }=\dfrac{\sin \theta }{\cos \theta }
We know, an identity that tan⁡θ\tan \theta equal to sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta } we can represent the expression sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta } as tan⁡θ\tan \theta which equals the expression of the right-hand side.
So, we get sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta .
Hence, it is proved.

Note: We can also use another identity to simplify and prove the result. After simplifying the expression of left-hand side into the form 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)} , we can use the identity sin⁡2θ{{\sin }^{2}}\theta equal to (1−cos⁡2θ)\left( 1-{{\cos }^{2}}\theta \right) and substitute it. Then, we will get the expression as sin⁡θ(2cos⁡2θ−1)cos⁡θ(2cos⁡2θ−1)\dfrac{\sin \theta \left( 2{{\cos }^{2}}\theta -1 \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}. The common terms can be cancelled off and again we will get on simplification, which is the same as right hand side and thus prove the result.