Question

How do you find the amplitude, period, phase shift for $y=\sin \left( x-\dfrac{\pi }{4} \right)$?

Answer 1

Consider the sine function $\sin x$and draw its graph to check its minimum and maximum value. This minimum or maximum value will be the amplitude of $\sin \left( x-\dfrac{\pi }{4} \right)$. Now, to find the time period of $\sin \left( x-\dfrac{\pi }{4} \right)$ first find the time period of $\sin x$ and assume it as ‘T’, then use the formula of time period of $\sin \left( ax+b \right)$ given as $T'=\dfrac{T}{\left| a \right|}$. Finally, determine the phase shift of $\sin \left( x-\dfrac{\pi }{4} \right)$ by comparing it with $\sin x$. Check the angle which is subtracted or added in x.

Complete step-by-step answer:
Here, we have been provided with the function $y=\sin \left( x-\dfrac{\pi }{4} \right)$ and we are asked to find its amplitude, time period and phase shift. But first we need to understand the meaning of these three terms. So, let us check them one – by – one.
(i) Amplitude: - Amplitude of a function is defined as the minimum and maximum value that a function can approach. Here, we have sine function. Let us draw the graph of $\sin x$ to check its minimum and maximum value.

From the above graph we can say that the range of $\sin x$ is [-1, 1]. It means sine function lies in the range [-1, 1].
$\Rightarrow -1\le \sin \left( x-\dfrac{\pi }{4} \right)\le 1$
Hence, the range of $\sin \left( x-\dfrac{\pi }{4} \right)$ will also be [-1, 1]. So, the above relation represents the amplitude of y.
(ii) Time period: - Time period of a function is defined as the value of x after which the value of the function starts repeating itself. It is also simply called the period and the function is known as periodic function.
From the above graph we can say that the period of $\sin x$ is $2\pi$. We know that if period of a function f (x) is T then the period of the function $f\left( ax+b \right)$ is given as: - $T'=\dfrac{T}{\left| a \right|}$. So, considering $f\left( x \right)=\sin x$ and $f\left( ax+b \right)=\sin \left( x-\dfrac{\pi }{4} \right)$ in the above question, we have,
$\Rightarrow a=1,b=\dfrac{-\pi }{4}$
So, the time period of $\sin \left( x-\dfrac{\pi }{4} \right)$ is given as: -

& \Rightarrow T'=\dfrac{2\pi }{1} \\\ & \Rightarrow T'=2\pi \\\ \end{aligned}$$ Hence, the time period of $$y=\sin \left( x-\dfrac{\pi }{4} \right)$$ is $$2\pi $$. (iii) Phase shift: - Phase shift of a function is defined as how far a function is shifted horizontally from the usual position. Now, for sine function given as $$\sin \left( x+a \right)$$ the phase shift is given as: - 1\. If ‘a’ is negative then phase shift is ‘a’ units to the right. 2\. If ‘a’ is positive then phase shift is ‘a’ units to the left. In the above question we have $$y=\sin \left( x-\dfrac{\pi }{4} \right)$$, so on comparing we have $$a=\dfrac{-\pi }{4}$$ which is negative. So, it is the $${{1}^{st}}$$ condition. Hence, the phase shift is $$\dfrac{\pi }{4}$$. **Note:** One may note that we have another type of phase shift known as vertical phase shift. That case arises if there is a constant multiplied to the function. It also changes the amplitude. In the above question it was 1 so it did not affect any relation. Note that phase shift has no effect on time period. Time period changes only when the coefficient of x is changed in the function. You must draw the graph of the given function as it will help us to understand the situation in a better way. You must remember the formula $$T'=\dfrac{T}{\left| a \right|}$$ as it will be used in chapters like integration.