Question

How do you find the exact value of \[{{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)\] ?

Answer 1

Hint:The given question is the inverse trigonometric expression and in order to solve this solve we have to use the properties of inverse trigonometric functions. We will have to first convert the given equation into the standard form i.e. in terms of cos and then by using its range, we will determine the exact value of \[{{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)\].

Formula used:
Trigonometric ratio table-

Angles(in degrees)	\[\sin \theta \]	\[\cos \theta \]
\[{{0}^{0}}\]	0	1
\[{{30}^{0}}\]	\[\dfrac{1}{2}\]	\[\dfrac{\sqrt{3}}{2}\]
\[{{45}^{0}}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{\sqrt{2}}\]
\[{{60}^{0}}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{2}\]
\[{{90}^{0}}\]	1	0

Range of inverse trigonometric functions:

Function	Range	Positive	Negative
\[{{\sin }^{-1}}\]	\[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\]	\[\theta \]	-\[\theta \]
\[{{\cos }^{-1}}\]	\[\left[ 0,\pi \right]\]	\[\theta \]	\[\pi -\theta \]
\[{{\tan }^{-1}}\]	\[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\]	\[\theta \]	-\[\theta \]

Complete step by step answer:
Here, we have given the function \[{{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)\] and we need to find out the exact value,
Let \[\theta ={{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)\]
By transferring \[{{\cos }^{-1}}\] to the left side of the equation, it will become cos function
We obtain,
\[\cos \theta =\left( \dfrac{\sqrt{2}}{2} \right)\]-------- (1)
By using trigonometric ratio table, we know that
\[\cos \dfrac{\pi }{4}=\dfrac{\sqrt{2}}{2}\]
Substituting this value in the equation (1), we obtain
\[\cos \theta =\cos \dfrac{\pi }{4}\]
As we know that the range of the principal value of \[{{\cos }^{-1}}\] is between \[[0,\pi ]\]
For positive values of the given function, we have the principal value \[\theta \] and for the negative values of the given function, we will have the principal value as \[\pi -\theta \].Here, \[\left( \dfrac{\sqrt{2}}{2} \right)\] is positive.Therefore, the principal value of the given function \[{{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)\] is \[\dfrac{\pi }{4}\].

Hence, the exact value will be \[\dfrac{\pi }{4}\].


Note: We should remember range and domain for trigonometric and inverse trigonometric functions. To solve these type of questions of inverse trigonometry, you always need to first convert it into standard form i.e. terms of trigonometric functions. The solution in which the absolute value of the angle is the least is called the principal value or the exact value.