Question

How do you evaluate $\cos \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)$ without a calculator?

Answer 1

Hint: For answering this question we have been asked to evaluate the value of $\cos \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)$ without using a calculator. From the basic concepts we know that the value of $\sin {{60}^{\circ }}$ is $\dfrac{\sqrt{3}}{2}$ and the value of $\cos {{60}^{\circ }}$ is $\dfrac{1}{2}$ .

Complete step by step solution:
Now considering from the question we need to evaluate the value of $\cos \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)$ .
From the basic concepts of trigonometry we know that the value of $\sin {{60}^{\circ }}$ is $\dfrac{\sqrt{3}}{2}$ and the value of $\cos {{60}^{\circ }}$ is $\dfrac{1}{2}$ .
Now we will substitute $\dfrac{\sqrt{3}}{2}$ with $\sin {{60}^{\circ }}$ , then we will have $\Rightarrow \cos \left( {{\sin }^{-1}}\left( \sin {{60}^{\circ }} \right) \right)$ .
From the basics of concepts we know that we have a formula in trigonometry mathematically given as ${{\sin }^{-1}}\left( \sin \theta \right)=\theta $ .
Now we will use this formula and further simplify the expression. After doing that we will have $\Rightarrow \cos \left( {{60}^{\circ }} \right)$ .
Now we will use the value of $\cos \left( {{60}^{\circ }} \right)$ and simplify it further.
After applying we will have $\Rightarrow \dfrac{1}{2}$ .

Therefore we can conclude that the value of the given expression $\cos \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)$ is given as $\dfrac{1}{2}$ .

Note: While answering questions of this type we should be sure with our concepts and calculations. This type of question can be answered in a short span of time and very few mistakes are possible. Similarly we can simplify any trigonometric expression for example if we consider the trigonometric expression $\sin \left( {{\cos }^{-1}}\left( \dfrac{1}{2} \right) \right)$ we can simplify this and evaluate it and then we will get $\sin \left( {{\cos }^{-1}}\left( \cos {{60}^{\circ }} \right) \right)=\sin {{60}^{\circ }}\Rightarrow \dfrac{\sqrt{3}}{2}$ because we know that the value of $\sin {{60}^{\circ }}$ is $\dfrac{\sqrt{3}}{2}$ and the value of $\cos {{60}^{\circ }}$ is $\dfrac{1}{2}$ .