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Question: To prove that \(3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)\), using a range \(x \...

To prove that 3sin⁡−1x=sin⁡−1(3x−4x3)3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right), using a range x∈[−12,12]x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right].

Explanation

Solution

In this problem, using the trigonometric function range table to define value and reverse trigonometric method for confirming equal on both sides. We're taking one side of it to solve this problem. And make the other side as the final solution.

Formula used: To put x=sin⁡θx = \sin \theta
Then sin⁡−1x=θ{\sin ^{ - 1}}x = \theta
The given trigonometric function of sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4{\sin ^3}x is derived to get (sin⁡−1(sin⁡x)=x)\left( {{{\sin }^{ - 1}}\left( {\sin x} \right) = x} \right)

Complete step-by-step answer:
Given that,
The value of L.H.S is 3sin⁡−1x3{\sin ^{ - 1}}x
the value of R.H.S is sin⁡−1(3x−4x3){\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)
Let take R.H.S values,
R.H.S
sin⁡−1(3x−4x3){\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)…………(1)(1)
To put x=sin⁡θx = \sin \theta
To substituting a value xx in given equation
We get,
⇒\Rightarrow sin⁡−1(3sin⁡θ−4(sin⁡θ)3){\sin ^{ - 1}}\left( {3\sin \theta - 4{{\left( {\sin \theta } \right)}^3}} \right)
Now we using a trigonometric formula, and the range of the function is 12⩽−4x3⩽−12\dfrac{1}{2} \leqslant - 4{x^3} \leqslant - \dfrac{1}{2}
We know that,
The trigonometric value of sin⁡−1(3sin⁡θ−4(sin⁡3θ)){\sin ^{ - 1}}\left( {3\sin \theta - 4\left( {{{\sin }^3}\theta } \right)} \right) the equation is given by
⇒\Rightarrow sin⁡−1(sin⁡3θ){\sin ^{ - 1}}\left( {\sin 3\theta } \right)………..(2)(2)
On simplifying the value,
Again, we use trigonometric formula in equation (2)(2)
We get the value of R.H.S is 3θ3\theta
By using the above formula, we apply for θ\theta value
To solving a given 3θ3\theta
Therefore,
⇒\Rightarrow 3θ=3\theta = 3sin⁡−1x3{\sin ^{ - 1}}x
Hence proved that the given L.H.S is equal to R.H.S
⇒\Rightarrow 3sin⁡−1x=sin⁡(3x−4x3)3{\sin ^{ - 1}}x = \sin \left( {3x - 4{x^3}} \right)

Note: Alternative method
Using the trigonometric range formula,
Therefore,
⇒\Rightarrow sin⁡3θ=3x−4x3\sin 3\theta = 3x - 4{x^{^3}}
On simplifying,
⇒\Rightarrow 3θ=sin⁡−1(3x−4x3)3\theta = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)
Using trigonometric formula,
x=sin⁡θx = \sin \theta
⇒\Rightarrow sin⁡−1x=θ{\sin ^{ - 1}}x = \theta
To apply the above formula,
We get,
⇒\Rightarrow 3sin⁡−1x=sin⁡−1(3x−4x3)3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)
Hence the solution is proved.