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Question

Question: How do you find the period of \[y=2\sin x\]?...

How do you find the period of y=2sinxy=2\sin x?

Explanation

Solution

The standard form of the sine function is y=asin(bx+c)+dy=a\sin \left( bx+c \right)+d, here a, b, c, and d \in Real numbers. For this equation, the amplitude is a\left| a \right|, period is 2πb\dfrac{2\pi }{b}, phase shift is cb\dfrac{-c}{b}, and vertical shift is d. By comparing the coefficients of the equation, we can find the a, b, c, d, and thus amplitude, period, and shifts can also be found.

Complete step by step answer:
We are given the cosine function y=2sinxy=2\sin x. We have to find its amplitude, period, and shifts. The standard form of the cosine function is y=asin(bx+c)+dy=a\sin \left( bx+c \right)+d. Comparing the given cosine equation with standard form we get, a=2,b=1,c=0,a=2,b=1,c=0, and d=0d=0.
We can find the period from the values of the coefficients as follows,
For a standard sine function y=asin(bx+c)+dy=a\sin \left( bx+c \right)+d, the period is 2πb\dfrac{2\pi }{b}, substituting the values we get, period of given function as 2πb=2π1=2π\dfrac{2\pi }{b}=\dfrac{2\pi }{1}=2\pi .

Hence, the period of the given sine function is 2π2\pi .

Note: We can also find the amplitude, horizontal/phase shift, and vertical shift from the value of coefficients as follows,
Amplitude of given equation is a=2=2\left| a \right|=\left| 2 \right|=2, horizontal/ phase shift of the equation is cb=01=0\dfrac{-c}{b}=\dfrac{0}{1}=0, and vertical shift of equation is d =0=0.
We can plot the graph of the given sine function using these values, as

A similar type of question can be asked for cosine functions also, for that we should know what the standard form of the cosine function is: y=acos(bx+c)+dy=a\cos \left( bx+c \right)+d. Same as for the sine function here also a, b, c, and d \in Real numbers. For this equation, the amplitude is a\left| a \right|, period is 2πb\dfrac{2\pi }{b}, phase shift is cb\dfrac{-c}{b}, and vertical shift is d.