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} $ .