
How do you find the amplitude, period, and phase shift for $y = cos\left( {\theta + 180^\circ } \right)$?
Answer
477.3k+ views
Hint: In the question, we have to determine the amplitude, period, and phase-shift for the given trigonometric expression. To determine amplitude, period, and phase-shift, we first determine the amplitude for the given trigonometric expression. The standard form of the equation is $y = A\cos \left( {kx + \psi } \right)$. Where A is the amplitude, k is the number of waves and the value of k is equal to $\dfrac{{2\pi }}{\lambda }$ , $\lambda $ is the wavelength that is called the period and $ - \dfrac{\psi }{k}$ is the phase-shift.
Complete step-by-step answer:
First, we will determine the amplitude for the given trigonometric expression.
In this question, the given expression is:
$ \Rightarrow y = cos\left( {\theta + 180^\circ } \right)$
Let us compare the above equation with the standard form of equation $y = A\cos \left( {kx + \psi } \right)$.
By comparing the equation, we get the value of A is 1, the value of k is 1, and the value of $\psi $ is $180^\circ $.
$ \Rightarrow A = 1$
$ \Rightarrow k = 1$
And $ \Rightarrow \psi = 180^\circ $.
Hence, the value of the amplitude A is 1.
As we know that $k = \dfrac{{2\pi }}{\lambda }$
Therefore,
$ \Rightarrow \lambda = \dfrac{{2\pi }}{k}$
Here, the value of k is 1.
$ \Rightarrow \lambda = \dfrac{{2\pi }}{1}$
That is equal to,
$ \Rightarrow \lambda = 2\pi $
Hence, the period $\lambda $is $2\pi $.
Now, we know the formula of phase -shift is $ - \dfrac{\psi }{k}$.
Put the value of $\psi = 180^\circ $ and $k = 1$.
Therefore,
$ \Rightarrow - \dfrac{{180^\circ }}{1}$
That is equal to,
$ \Rightarrow - 180^\circ $
Hence, the value of the amplitude is 1, the value of period is $2\pi $, and the value of phase-shift is $ - 180^\circ $.
Note:
By comparing the given expression with the standard form, we can get the value of amplitude and the value of k. To obtain the value of the period, we have to put the value of k in the formula of the period that is $\lambda = \dfrac{{2\pi }}{k}$. We can also get the value of phase-shift by putting the value of k and $\psi $ in the formula of phase-shift that is $ - \dfrac{\psi }{k}$.
Complete step-by-step answer:
First, we will determine the amplitude for the given trigonometric expression.
In this question, the given expression is:
$ \Rightarrow y = cos\left( {\theta + 180^\circ } \right)$
Let us compare the above equation with the standard form of equation $y = A\cos \left( {kx + \psi } \right)$.
By comparing the equation, we get the value of A is 1, the value of k is 1, and the value of $\psi $ is $180^\circ $.
$ \Rightarrow A = 1$
$ \Rightarrow k = 1$
And $ \Rightarrow \psi = 180^\circ $.
Hence, the value of the amplitude A is 1.
As we know that $k = \dfrac{{2\pi }}{\lambda }$
Therefore,
$ \Rightarrow \lambda = \dfrac{{2\pi }}{k}$
Here, the value of k is 1.
$ \Rightarrow \lambda = \dfrac{{2\pi }}{1}$
That is equal to,
$ \Rightarrow \lambda = 2\pi $
Hence, the period $\lambda $is $2\pi $.
Now, we know the formula of phase -shift is $ - \dfrac{\psi }{k}$.
Put the value of $\psi = 180^\circ $ and $k = 1$.
Therefore,
$ \Rightarrow - \dfrac{{180^\circ }}{1}$
That is equal to,
$ \Rightarrow - 180^\circ $
Hence, the value of the amplitude is 1, the value of period is $2\pi $, and the value of phase-shift is $ - 180^\circ $.
Note:
By comparing the given expression with the standard form, we can get the value of amplitude and the value of k. To obtain the value of the period, we have to put the value of k in the formula of the period that is $\lambda = \dfrac{{2\pi }}{k}$. We can also get the value of phase-shift by putting the value of k and $\psi $ in the formula of phase-shift that is $ - \dfrac{\psi }{k}$.
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