Question

Prove the following:
$3{{\sin }^{-1}}x={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)$ , $x\in \left[ -\dfrac{1}{2},\dfrac{1}{2} \right]$.

Answer 1

Hint: At first write the identity \[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta \] . Then take $\sin \theta $ as x. So it can written as $\sin 3\theta =3x-4{{x}^{3}}$ then use the fact $\theta $ is ${{\sin }^{-1}}x$

Complete step-by-step answer:
In the question we are given the equation $3{{\sin }^{-1}}x={{\sin }^{-1}}\left( 3x-4{{x}^{3}}

\right)$.If x belongs to the range $\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]$ .

Before proceeding let us know what are inverse trigonometric functions. Inverse trigonometric functions are functions that are inverse functions of trigonometric functions. Specifically they are inverses of sine, cosine, tangent, cotangent, secant, cosecant functions, and are used to obtain an angle from any of the angles' trigonometric ratios.

These are certain notations which are used. Some of the most common notation is using arc $\sin \left( x \right)$ , arc $\cos \left( x \right)$ , arc $\tan \left( x \right)$ instead of ${{\sin }^{-1}}\left( x \right)$ , ${{\cos }^{-1}}\left( x \right)$ and ${{\tan }^{-1}}\left( x \right)$ . These arise from geometric relationships. When measuring in radius, an angle $\theta $ radius will correspond to an arc whose length is r $\theta $ , where r is radius of circle. Thus in the unit circle, “the arc whose cosine is x” is the same as “the angle whose cosine is x”, and the length of the arc of the circle in radii is the same as the measurement of angle in radius.

Before proving first check that ${{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)$ exist or not by proving that the range of $3x-4{{x}^{3}}$ varies between -1 and 1 or not.

So, we are given that $x\in \left[ -\dfrac{1}{2},\dfrac{1}{2} \right]$ which can be written as –

$-\dfrac{1}{2}\le x\le \dfrac{1}{2}$ ………………………. (i)

We can multiply the whole equation (i) by 3. So we get as –

$-\dfrac{3}{2}\le 3x\le \dfrac{3}{2}$ ………………………. (ii)

Now, we will cube to all the sides of equation we get –

$-\dfrac{1}{8}\le {{x}^{3}}\le \dfrac{1}{8}$ .

Now, we will multiply by $\left( -4 \right)$ to all the sides. We can write it as –

$\dfrac{1}{2}\le -4{{x}^{3}}\le -\dfrac{1}{2}$ …………… (iii)

Now, we will add equations (ii) and (iii). So we get –

$-\dfrac{3}{2}+\dfrac{1}{2}\le 3x-4{{x}^{3}}\le \dfrac{3}{2}-\dfrac{1}{2}$ .

So, on simplification we get –

$-1\le 3x-4{{x}^{3}}\le 1$.

Hence, the value of $3x-4{{x}^{3}}$ lies between -1 and 1. So its value exists.

Now, let’s suppose $x=\sin \theta $ or $\theta ={{\sin }^{-1}}x$.

We know a formula that, $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta $ .

Now instead of $\sin \theta $ we can put x so we get –

$\sin 3\theta =3x-4{{x}^{3}}$.

Hence it can be written as –

$3\theta ={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)$ .

We also know that $\theta ={{\sin }^{-1}}x$ so we can rewrite equation as –

$3{{\sin }^{-1}}x={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)$ .

Hence, proved.


Note: The value of ${{\sin }^{-1}}\left( x \right)$ lies between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$ so x will always lies between -1 and 1. To prove that we can directly put x as $\sin \theta $ then we can use identity $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta $ and get the answer.