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Question: Prove \[3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3}),x \in [ - \dfrac{1}{2},\dfrac{1}{2}]\]...

Prove 3sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12]3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3}),x \in [ - \dfrac{1}{2},\dfrac{1}{2}]

Explanation

Solution

We are given the range of the variable xx. Here we can introduce another variable θ\theta such that x=sin⁡θx = \sin \theta . We know that sine function is continuous and invertible in the interval [−π,π][ - \pi ,\pi ]. So we can consider θ=sin⁡−1x\theta = {\sin ^{ - 1}}x in the interval [−12,12][ - \dfrac{1}{2},\dfrac{1}{2}]. Substituting x=sin⁡θx = \sin \theta , we can easily prove the result using known trigonometric relations.

Formula used:
We have the trigonometric result;
For any angle θ\theta , sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta

Complete step-by-step answer:
Given that 3sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12]3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3}),x \in [ - \dfrac{1}{2},\dfrac{1}{2}]
We have x∈[−12,12]x \in [ - \dfrac{1}{2},\dfrac{1}{2}].
Let x=sin⁡θx = \sin \theta
Then sin⁡−1x=sin⁡−1(sin⁡θ)=θ{\sin ^{ - 1}}x = {\sin ^{ - 1}}(\sin \theta ) = \theta .
Since sin⁡−1x{\sin ^{ - 1}}x is a continuous function in the interval [−12,12][ - \dfrac{1}{2},\dfrac{1}{2}], then there exist some θ\theta for every x∈[−12,12]x \in [ - \dfrac{1}{2},\dfrac{1}{2}].
So we have x=sin⁡θx = \sin \theta
Now to prove 3sin⁡−1x=sin⁡−1(3x−4x3)3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3}), start from the right hand side.
Substituting for xx we have,
sin⁡−1(3x−4x3)=sin⁡−1(3sin⁡θ−4sin⁡3θ){\sin ^{ - 1}}(3x - 4{x^3}) = {\sin ^{ - 1}}(3\sin \theta - 4{\sin ^3}\theta )
We know that sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta
Substituting this result in the above equation we get,
sin⁡−1(3x−4x3)=sin⁡−1(sin⁡3θ){\sin ^{ - 1}}(3x - 4{x^3}) = {\sin ^{ - 1}}(\sin 3\theta )
⇒sin⁡−1(3x−4x3)=3θ\Rightarrow {\sin ^{ - 1}}(3x - 4{x^3}) = 3\theta
Now, x=sin⁡θ⇒θ=sin⁡−1xx = \sin \theta \Rightarrow \theta = {\sin ^{ - 1}}x
Substituting this we get,
⇒sin⁡−1(3x−4x3)=3sin⁡−1x\Rightarrow {\sin ^{ - 1}}(3x - 4{x^3}) = 3{\sin ^{ - 1}}x
Thus we had reached the left hand side.
Therefore we have,
3sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12]3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3}),x \in [ - \dfrac{1}{2},\dfrac{1}{2}]
Hence we have proved the result.

Additional information:
We can also make the substitution cos⁡θ\cos \theta for similar problems. Cosine function also has these properties used here. In the case of cosine, the trigonometric result is different.
cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta
So if the expression inside the bracket of the right hand side was like these we use the substitution x=cos⁡θx = \cos \theta .

Note: For proving a result we have to prove it generally. Here we considered x=sin⁡θx = \sin \theta . This is sufficient since we can see that there exists a θ\theta corresponding to every x∈[−12,12]x \in [ - \dfrac{1}{2},\dfrac{1}{2}] satisfying this. We know that, sin⁡(−30∘)=−12\sin ( - 30^\circ ) = - \dfrac{1}{2} and sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2}. So for x=12x = \dfrac{1}{2}, the value of θ\theta is 30∘30^\circ . Here the continuity can be understood by the continuity of the sine graph.