Question

If $x = {\sin ^{ - 1}}\left( {3t - 4{t^3}} \right)$ and $y = {\cos ^{ - 1}}\left( {\sqrt {1 - {t^2}} } \right)$ then $\dfrac{{dy}}{{dx}}$ is equal to?

Answer 1

To solve this type of questions, the best approach is to always remember the standard trigonometric identities. Notice here that the expression given for $x = {\sin ^{ - 1}}\left( {3t - 4{t^3}} \right)$ can be simplified using the standard formula of $\sin 3\theta$ and the expression given $y = {\cos ^{ - 1}}\left( {\sqrt {1 - {t^2}} } \right)$ can be reduced to a simpler form by using standard identity ${\sin ^2}\theta + {\cos ^{^2}}\theta = 1$ . So basic algebraic rules and trigonometric identities
have to be kept in mind while solving the given problem.

Complete step by step answer:
To simplify the question, let us assume that $t = \sin \theta$ . Now, put the value of $t$ in the expression for
$x$ , we get;
$\Rightarrow x = {\sin ^{ - 1}}\left( {3t - 4{t^3}} \right)$
$\Rightarrow x = \left( {3\sin \theta - 4{{\sin }^3}\theta } \right)$
By identity, $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$ the above expression can be reduced to;
$\Rightarrow x = {\sin ^{ - 1}}\left( {\sin 3\theta } \right)$
By identity, ${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta$ therefore we get;
$\Rightarrow x = 3\theta$
Now, let us try to simplify the given expression for $y$ by replacing $t = \sin \theta$ , we get;
$\Rightarrow y = {\cos ^{ - 1}}\left( {\sqrt {1 - {t^2}} } \right)$
$\Rightarrow y = {\cos ^{ - 1}}\left( {\sqrt {1 - {{\sin }^2}\theta } } \right)$
By identity, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we know $\sqrt {1 - {{\sin }^2}\theta } = \cos \theta$ , we get;
$\Rightarrow y = {\cos ^{ - 1}}\left( {\cos \theta } \right)$
$\Rightarrow y = \theta$
Now, we will calculate the derivative of $x$ and $y$;
Differentiating $x$ with respect to (w.r.t.) $\theta$ , we get;
$\because x = 3\theta$
$\therefore \dfrac{{dx}}{{d\theta }} = 3\dfrac{d}{{d\theta }}\left( \theta \right)$
By the differentiation formula, $\dfrac{d}{{dx}}\left( x \right) = 1$, we get;
$\Rightarrow \dfrac{{dx}}{{d\theta }} = 3$ $......\left( 1 \right)$
Now Differentiating $y$ with respect to (w.r.t.) $\theta$ , we get;
$\because y = \theta$
$\therefore \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( \theta \right)$
$\Rightarrow \dfrac{{dy}}{{d\theta }} = 1$ $......\left( 2 \right)$
Now, according to the given question we have to calculate $\dfrac{{dy}}{{dx}}$ therefore dividing equation $\left( 2 \right)$ by equation $\left( 1 \right)$, we get;
$\Rightarrow \dfrac{{\left( {\dfrac{{dy}}{{d\theta }}} \right)}}{{\left( {\dfrac{{dx}}{{d\theta }}} \right)}} = \dfrac{1}{3}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{3}$
Therefore, the value of $\dfrac{{dy}}{{dx}}$ is $\dfrac{1}{3}$.
This question can also be solved like this:
Let $t = \sin \theta$
So, $\theta = {\sin ^{ - 1}}\left( t \right)$
By identity, $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$
$\Rightarrow x = {\sin ^{ - 1}}\left( {3t - 4{t^3}} \right)$
$\Rightarrow x = {\sin ^{ - 1}}\sin \left( {3\theta } \right)$
$\Rightarrow x = 3\theta$
$\Rightarrow x = 3{\sin ^{ - 1}}t$ $\left( {\because \theta = {{\sin }^{ - 1}}t} \right)$
By identity , ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\Rightarrow \cos \theta = \pm \sqrt {1 - {{\sin }^2}\theta }$
$\Rightarrow y = {\cos ^{ - 1}}\left( {\sqrt {1 - {t^2}} } \right)$
$\Rightarrow y = {\cos ^{ - 1}}\cos \left( \theta \right)$
$\Rightarrow y = \theta$
$\Rightarrow y = {\sin ^{ - 1}}t$
To calculate $\dfrac{{dy}}{{dx}}$ , we should calculate $\dfrac{{dy}}{{dt}}$ and $\dfrac{{dx}}{{dt}}$.
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {{{\sin }^{ - 1}}t} \right)$
By standard formula; $\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Therefore $\dfrac{{dy}}{{dt}} = \dfrac{1}{{\sqrt {1 - {t^2}} }}$
Now, we will calculate $\dfrac{{dx}}{{dt}}$
$\Rightarrow \dfrac{{dx}}{{dt}} = 3\dfrac{d}{{dt}}\left( {{{\sin }^{ - 1}}t} \right)$
Therefore $\dfrac{{dx}}{{dt}} = \dfrac{3}{{\sqrt {1 - {t^2}} }}$
Now , calculating $\dfrac{{dy}}{{dx}}$ by putting the values of $\dfrac{{dy}}{{dt}}$ and $\dfrac{{dx}}{{dt}}$ ;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{1}{{\sqrt {1 - {t^2}} }}}}{{\dfrac{3}{{\sqrt {1 - {t^2}} }}}}$
On further calculation;
Therefore the value of $\dfrac{{dy}}{{dx}}$ is given as ;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{3}$

Note:
The given question deals with basic simplification of trigonometric functions by using some of the standard trigonometric identities such as; $\sin 3 \theta= 3 \sin \theta - 4 \sin^3 \theta$ and $\sin^2 \theta + \cos^2 \theta = 1$.
Besides this, simple algebraic rules and identities are also of significant use in such types of problems.