Question

How do you evaluate $ {\cos ^{ - 1}}\left( 1 \right) $ ?

Answer 1

Hint: In this question we will evaluate $ {\cos ^{ - 1}}\left( 1 \right) $ . We will consider the given as $ x $ and compare the value with the trigonometric table and determine the value of $ x $ and we will construct an equation based on the occurrence of the value in the trigonometric table.

Complete step-by-step answer:
Here, we will evaluate $ {\cos ^{ - 1}}\left( 1 \right) $ .
Now, let us consider $ {\cos ^{ - 1}}\left( 1 \right) = x $ $ \to \left( 1 \right) $
Then, $ \cos x = 1 $
Now, we know from the trigonometric table that $ \cos 0^\circ = 1 $ .
The domain of inverse cosine function is $ \left[ { - 1,1} \right] $ and the range is $ \left[ {0,\pi } \right] $ . This means a positive value will yield a first quadrant value and a negative value will yield a second quadrant angle.
Now, let us substitute the value of $ \cos 0^\circ = 1 $ in $ \cos x = 1 $ ,
 $ \cos x = \cos 0^\circ $
Therefore, $ x = 0 $
Since, $ 1 $ represents the maximum value of the cosine function. It happens at $ 0 $ , $ 2\pi $ , $ 4\pi $ , $ 6\pi $ , etc.
Thus, we can write $ x = 0 + 2\pi k $ .
Hence, from the equation $ \left( 1 \right) $ , $ {\cos ^{ - 1}}x = 0 + 2\pi k $ where for any integer $ k $
So, the correct answer is “ $ {\cos ^{ - 1}}x = 0 + 2\pi k $”.

Note: In this question it is important to note here that inverse trigonometric functions are the inverse of the basic trigonometric functions. That contains sine, cosine, tangent, cosecant, secant and cotangent. They are also known as arcus functions, ant trigonometric functions or cyclometric functions. They perform opposite operations of the trigonometric functions. There are six important inverse trigonometric functions; they are arcsine, arccosine, arctangent, arccosecant, arcsecant and arccotangent.
The first quadrant is common to all inverse functions. Third quadrant is not used in inverse functions. Fourth quadrant is used in clockwise direction i.e., $ - \dfrac{\pi }{2} \leqslant y \leqslant 0 $ .
 $ {\cos ^{ - 1}}x $ is bounded in $ \left[ {0,\pi } \right] $ . $ {\cos ^{ - 1}}x $ is a neither odd nor even function. In its domain, $ {\cos ^{ - 1}}x $ is a decreasing function. $ {\cos ^{ - 1}}x $ attains its maximum value $ \pi $ at $ x = - 1 $ while its minimum value is $ 0 $ which occurs at $ x = 1 $ .