Question

How do you find the exact value of $\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)$ ?

Answer 1

Hint : We explain the function $\arcsin \left( x \right)$ . We express the inverse function of tan in the form of $\arcsin \left( x \right)={{\sin }^{-1}}x$ . We draw the graph of $\arcsin \left( x \right)$ and the line $x=\dfrac{1}{3}$ to find the intersection point. Thereafter we take the cos ratio of that angle to find the solution.

Complete step-by-step answer :
The given expression is the inverse function of trigonometric ratio sin.
The arcus function represents the angle which on ratio tan gives the value.
So, $\arcsin \left( x \right)={{\sin }^{-1}}x$ . If $\arcsin \left( x \right)=\alpha$ then we can say $\sin \alpha =x$ .
Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of $2\pi$ .
The general solution for that value where $\sin \alpha =x$ will be $n\pi +{{\left( -1 \right)}^{n}}\alpha ,n\in \mathbb{Z}$ .
But for $\arcsin \left( x \right)$ , we won’t find the general solution. We use the principal value. For ratios sin we have $-\dfrac{\pi }{2}\le \arcsin \left( x \right)\le \dfrac{\pi }{2}$ .
The graph of the function is
We now place the value of $x=\dfrac{1}{3}$ in the function of $\arcsin \left( x \right)$ .
Let the angle be $\theta$ for which $\arcsin \left( \dfrac{1}{3} \right)=\theta$ . This gives $\sin \theta =\dfrac{1}{3}$ .
The value of $\theta$ for which $\sin \theta$ is $\dfrac{1}{3}$ is 19.47 degree..
Now we take $\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\cos \left( {{19.47}^{\circ }} \right)=\dfrac{\sqrt{8}}{3}$ .
Therefore, the value of $\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)$ is $\dfrac{\sqrt{8}}{3}$ .
So, the correct answer is “$\dfrac{\sqrt{8}}{3}$ ”.

Note : We can also apply the trigonometric image form to get the value of $\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)$ .
It’s given that $\sin \theta =\dfrac{1}{3}$ and we need to find $\cos \theta$ . We know $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$ .
Putting the values, we get $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{\left( \dfrac{1}{3} \right)}^{2}}}=\dfrac{\sqrt{8}}{3}$ .