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Question

Question: How do you evaluate \(\arccos ( - 1)\) without a calculator?...

How do you evaluate arccos(1)\arccos ( - 1) without a calculator?

Explanation

Solution

We have the arccos(1)\arccos ( - 1) cosine of the term and we have to simplify it; we will use the domain and range of the cosine function and find the required solution.

Complete step-by-step solution:
We have the given trigonometric function given to us as: arccos(1)\arccos ( - 1)
The function arccos(x)\arccos (x) is the inverse function to the cos(x)\cos (x) function, therefore in this question we have to find the angle which the function arccos(1)\arccos ( - 1) will give to us.
To find the required solution, we first have to find out the domain and range of the function arccos(x)\arccos (x)
We know the restrictions to the function are as follows:
The domain of the function is [1,1][ - 1,1] and the range of the function is [0,π][0,\pi ]
Therefore, the only angle which has a cos(1)=π\cos ( - 1) = \pi because we know that cos(0)\cos (0) is 11, now since we have the function as negative 11, it will be in the second quadrant and the opposite value of 11.

Therefore arccos(1)=π\arccos ( - 1) = \pi

Note: Basic trigonometric formulas should be remembered to solve these types of sums.
The inverse trigonometric function of cosx\cos x which is arccos(x)\arccos (x) used in this sum
For example, if cosx=a\cos x = a then x=arccos(x)x = \arccos (x)
In some questions the inverse function is written as: cos1x{\cos ^{ - 1}}x, which is the function as arccos(x)\arccos (x). It does the same work as the function arccos(x)\arccos (x) does.
And x=arccos(cos(x))x = \arccos (\cos (x)) is a property of the inverse function.
There also exists inverse functions for the other trigonometric relations such as sin\sin and tan\tan .
The inverse function is used to find the angle xx from the value of the trigonometric function.
It is to be remembered that whenever there is a question on trigonometric functions, the question should be shifted in the simpler form of cos\cos and tan\tan .