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Question

Question: How do you find the amplitude and period of \(y = \dfrac{1}{2}\sin \theta ?\)...

How do you find the amplitude and period of y=12sinθ?y = \dfrac{1}{2}\sin \theta ?

Explanation

Solution

Amplitude is the highest value of a function in its one complete cycle or one complete period.
Period is the smallest length that repeats itself in a repeating or periodic function. And since, all the trigonometric functions are periodic, hence sin\sin function is also a periodic function.

Complete step by step answer:
As we already know that the amplitude of function means the highest possible value of that function. So do you know what’s the highest value of a sin\sin function?
Let us find out the highest value of a sin\sin function with the help of its graph.


Now, from the graph we know the highest value of a sin\sin function is 11, therefore required amplitude will be the highest value of yy in the equation
y=12sinθy = \dfrac{1}{2}\sin \theta
Here in the equation 12\dfrac{1}{2} is constant, so the only variable which can affect the value of yy is sinθ\sin \theta
And from the graph, we know the highest value of sinθ\sin \theta , which is equal to 11
So, substituting the highest value of sinθ\sin \theta which is 11 in the above equation we get,
y=12sinθ y=12×1 y=12  \Rightarrow y = \dfrac{1}{2}\sin \theta \\\ \Rightarrow y = \dfrac{1}{2} \times 1 \\\ \Rightarrow y = \dfrac{1}{2} \\\
Therefore the amplitude of the function y=12sinθy = \dfrac{1}{2}\sin \theta is 12\dfrac{1}{2}
Now, coming to the period as from the above graph we get to know that the period of sin\sin function is 2π2\pi or 360{360^ \circ } (in degrees)
But we have to find the period of 12sinθ\dfrac{1}{2}\sin \theta ,
Here, we can see that 12\dfrac{1}{2} is multiplied to the outcomes or yy - values of the function.
Therefore it will not affect the period until and unless the angle part which is θ\theta being multiplied or divided by it.

Therefore, the required period is 2π2\pi and amplitude is 12\dfrac{1}{2}

Note: When tackling this type of more questions then here is the general formula to find amplitude and period of a sine function. If sin\sin function is written as asinbθa\sin b\theta , then the amplitude and period is given by Amplitude =a = a and period =2πb = \dfrac{{2\pi }}{b}