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Question: How do you find the amplitude and period of arc cot?...

How do you find the amplitude and period of arc cot?

Explanation

Solution

Now to find the amplitude and period of the function we will first write the function in general form acot1(bxc)a{{\cot }^{-1}}\left( bx-c \right) now a is nothing but amplitude of the function and the period of the function is given by 2πb\dfrac{2\pi }{\left| b \right|} . Hence we can easily find the period and amplitude of the function.

Complete step by step solution:
Now we are given an inverse trigonometric function arc cot. We want to find the amplitude and period of the given function.
Now let us first understand the concept of amplitude and period.
Amplitude of any function is the maximum height of the function measured from the axis of the function.
Time period of the function is the length of the smallest interval after which a periodic function repeats its values.
Let us say we have sin(x)=sin(2π+x)\sin \left( x \right)=\sin \left( 2\pi +x \right) .
Hence we have 2π2\pi is the period of the function.
Now for a function of the form acot1(b(xc))a{{\cot }^{-1}}\left( b\left( x-c \right) \right) we have the amplitude of the function is equal to a and the time period of the function is equal to 2πb\dfrac{2\pi }{b} .
Now consider the given function cot1x{{\cot }^{-1}}x .
We can write this function as 1cot1(1(x0))1{{\cot }^{-1}}\left( 1\left( x-0 \right) \right)
Hence we have a = 1 and b = 1.
Hence the amplitude of the function is 1 and the period of the function is 2π2\pi .

Note: Now note that for a periodic function is 2π2\pi is the period then the function will repeat its value after every 2nπ2n\pi where n is any natural number. Hence note that while taking period we take smallest interval after which the function repeats.