Question

How do you solve and find the value of $\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{2}{9}} \right)} \right)$?

Answer 1

Here, in the given question, in need to solve and find the value of $\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{2}{9}} \right)} \right)$. As we can see here we are given a trigonometric function and an inverse trigonometric function. Trigonometric functions can be simply defined as functions of an angle of a triangle and inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions, they are also termed as arcus functions. As we know there is a relation between trigonometric function and an inverse trigonometric function which is given as; $\cos \left( {{{\cos }^{ - 1}}\theta } \right) = \theta$ so, by using this formula we will find the value of $\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{2}{9}} \right)} \right)$.

Formula used:
$\cos \left( {{{\cos }^{ - 1}}\theta } \right) = \theta$ for all $\theta \in \left[ { - 1,1} \right]$

Complete step by step answer:
We have, $\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{2}{9}} \right)} \right)$
As we know $\cos \left( {{{\cos }^{ - 1}}\theta } \right) = \theta$. Therefore, we get
$\Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{2}{9}} \right)} \right) = \dfrac{2}{9}$
We can write answers in terms of decimals also. Therefore, we get
$\Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{2}{9}} \right)} \right) = 0.222$

Therefore, the value of $\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{2}{9}} \right)} \right)$ is $\dfrac{2}{9}$.

Note: Remember that for any function $f$ and inverse of it i.e., ${f^{ - 1}}$, $f\left( {{f^{ - 1}}\left( x \right)} \right) = x$ and ${f^{ - 1}}\left( {f\left( x \right)} \right) = x$ are same. Remember that inverse trigonometric functions do the opposite of the regular trigonometric functions. For example: inverse $\cos$ i.e., ${\cos ^{ - 1}}$ does the opposite of $\cos$. The expression ${\cos ^{ - 1}}x$ is not the same as $\dfrac{1}{{\cos x}}$. In other words, the $- 1$ is not an exponent. Instead, it simply means inverse function.