Question

How do you find the period of $y = \cot (x - \dfrac{\pi }{2})$ ?

Answer 1

Hint: We are given a function of cotangent whose period we have to find so we have to first convert the given function to its standard form and then compare it with the standard equation for finding out the period of the function. The standard form of a trigonometric function is $f(x) = a.trig.(bx + c) + d$

Complete step-by-step solution:
Comparing $y = \cot (x - \dfrac{\pi }{2})$ with the standard form $y = a\cot (bx + c) + d$ , we get –
$a = 1,\,b = 1,\,C = - \dfrac{\pi }{2}\,and\,D = 0$
Thus, the given sine function has peak values at 1 and -1, that is, it oscillates between 1 and -1 and the given cotangent function completes one oscillation between 0 and $\pi $ .
We have to find out the period of the function. Period of this cotangent function is –
$
  p = \dfrac{{interval}}{b} \\
   \Rightarrow p = \dfrac{\pi }{1} = \pi \\
 $
That is after every $\pi $ radians the given function repeats the oscillation

Hence, the period of the function $y = \cot (x - \dfrac{\pi }{2})$ is equal to $\pi $.

Note: We know that the general form of the cosine function is $y = a\cot (bx + c) + d$ where
- $a$ is the amplitude, the peak values of a function are known as its amplitude.
- $b$ is the frequency, the number of oscillations that a function does in a fixed interval is known as its frequency.
- $c$ and $d$ tell us the horizontal and vertical shift of a function respectively. There is no horizontal or vertical shift in the given function as the value of C and D is zero for the given function.