How do you find the amplitude, period, and phase shift for $y = cos\left( {\theta + 180^\circ } \right)$?
Answer
612.6k+ 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}$.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

