Question

How do you solve and find the value of $\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)$ ?

Answer 1

Hint: We are asked to solve and find the value of $\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)$ , we start our solution by considering ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ as $\theta $ , then of the that we use $\sin \left( 2\theta \right)=2\sin \theta \cos \theta $ identity, after that we will learn how the sin and ${{\sin }^{-1}}$ are connected to each other, which we use further to solve the problem, we will also use ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to change the $\theta $ to cos theta or sin theta into one another.

Complete step by step solution:
We are given a function as $\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)$ , we have to evaluate the value of this function. To do so, we will learn that trigonometric functions are related to its inverses and we can also convert one trigonometric function to another.
Now as we have $\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)$ .
So, we consider ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ as $\theta $ then our equation become –
$\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\sin \left( 2\theta \right)$
Now, we know that –
$\sin 2\theta =2\sin \theta \cos \theta $
So, we get –
$\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=2\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right).\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)$ …………………………………. (1)
Now we learn about how the ratio and its inverse are connected to each other.
For sin we have that $\sin \left( {{\sin }^{-1}}\theta \right)=\theta $
And similarly for cos, we have $\cos \left( {{\cos }^{-1}}\theta \right)=\theta $ in equation (1), we have $\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)$ .
So, using above identity we get –
$\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\dfrac{1}{2}$
For $\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)$ we convert ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ into ${{\cos }^{-1}}$ .
Now, as we know, $\theta ={{\sin }^{-1}}\dfrac{1}{2}$ so, $\sin \theta =\dfrac{1}{2}$
We know ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Hence $\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }$ .
Hence, we get –
$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }$
As ${{\sin }^{2}}\theta =\dfrac{1}{2}$ so,
$\cos \theta =\pm \sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}$
By simplifying, we get –
$\cos \theta =\pm \sqrt{\dfrac{3}{4}}$ (as $1-{{\left( \dfrac{1}{2} \right)}^{2}}=1-\dfrac{1}{4}=\dfrac{3}{4}$ )
So, $\cos \theta =\pm \dfrac{\sqrt{3}}{2}$
Hence, $\theta ={{\cos }^{-1}}\left( \pm \dfrac{\sqrt{3}}{2} \right)$
So, we get –
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \pm \dfrac{\sqrt{3}}{2} \right)$
Thus $\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\cos \left( {{\cos }^{-1}}\left( \pm \dfrac{\sqrt{3}}{2} \right) \right)$
We get –
$=\pm \dfrac{\sqrt{3}}{2}$
So, we get two $\theta $ values
Case (I) When $\cos .{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=+\dfrac{\sqrt{3}}{2}$
Then as ${{\sin }^{-1}}\left( \sin \left( \dfrac{1}{2} \right) \right)=\dfrac{1}{2}$ and $\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\dfrac{\sqrt{3}}{2}$
So using these in equation (1) we get –
$\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=2\times \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}$
$=\dfrac{\sqrt{3}}{2}$
Case (II) when $\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=-\dfrac{\sqrt{3}}{2}$ as ${{\sin }^{-1}}\left( \sin \left( \dfrac{1}{2} \right) \right)=\dfrac{1}{2}$ and $\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=-\dfrac{\sqrt{3}}{2}$
We get by using these in equation (1)
$\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=2\times \dfrac{1}{2}\times \dfrac{-\sqrt{3}}{2}$
$=-\dfrac{\sqrt{3}}{2}$
Hence, we get –
$\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\pm \dfrac{\sqrt{3}}{2}$.

Note: While solving such problems we should be very careful with identity like $\sin 2x\ne 2\sin x$ or $\cos 2\theta \ne 2\cos \theta \sin \theta $ .
We should not mix or use appropriate identity; also we should always cross check solutions so that change at error will get eliminated.
While solving fractions we always report answers in the simplest form so if something is common we need to cancel it always.