Question

How do you find the exact value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)?$

Answer 1

Hint: The given function is the inverse trigonometry function; it's simply defined as the inverse function of the basic trigonometric function. This inverse function is used to find any angle of the trigonometric function.
In the question, the trigonometric function is inverse of the $\cos ine$function we can rewrite it as $\arccos ine$and it is expressed as $y={{\cos }^{-1}}\theta $.
The domain and the range of $y={{\cos }^{-1}}\theta $ is $-1\le \theta \le 1$ and $0\le y\le \pi $
The value for$\cos \theta $ of the following angle $0,{{30}^{0}},{{45}^{0}},{{60}^{0}},{{90}^{0}}$is

$\theta $	${{0}^{0}}$	${{30}^{0}}$	${{45}^{0}}$	${{60}^{0}}$	${{90}^{0}}$
$\cos $	$1$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$	$0$

Complete step by step solution:
Let us assume that $\theta ={{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$
And we know that the $\cos \dfrac{\pi }{6}=\left( \dfrac{\sqrt{3}}{2} \right)$
Thus, the above terms can be written as
$\Rightarrow \theta ={{\cos }^{-1}}\left( \cos \dfrac{\pi }{6} \right)$
$\Rightarrow \theta =\dfrac{\pi }{6}$
Therefore as we can see in the above the value of $\theta $is
$\Rightarrow \dfrac{\pi }{6}$
By using the ASTC rule we can see that $\cos ine$ is positive in the first quadrant and also in the fourth quadrant.
In the first quadrant, it will be $\dfrac{\pi }{6}$ and
In the fourth quadrant, it will be as $2\pi -\theta $
$\Rightarrow 2\pi -\dfrac{\pi }{6}$
$\Rightarrow \dfrac{12\pi -\pi }{6}$
$\Rightarrow \dfrac{11\pi }{6}$

Hence the exact value of the ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ will be $\dfrac{\pi }{6}$.

Note:
The inverse function of trigonometric is also known as “arcus trigonometric function” or “ant trigonometric function”.
The inverse function of trigonometric helps us to find out the unknown values of angles through the trigonometric ratio.
There are six trigonometric functions for each trigonometric ratio and the inverse of all those six functions as follow:
${{\sin }^{-1}}$or Arcsine
$co{{\operatorname{s}}^{-1}}$or Arccosine
${{\tan }^{-1}}$ or Arctangent
${{\csc }^{-1}}$or Arcosecant
${{\sec }^{-1}}$or Arcsecant
${{\cot }^{-1}}$or Arccotangent