Question

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

Answer 1

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

Formula used:
We have the trigonometric result;
For any angle $\theta $, \[\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \]

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

Additional information:
We can also make the substitution $\cos \theta $ for similar problems. Cosine function also has these properties used here. In the case of cosine, the trigonometric result is different.
$\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 \theta $.

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