Question

State the amplitude, period and phase shift of the function $y=\cos \left(\theta -45°\right)$ and then graph the function.

Short answer

The amplitude of $y=\cos \left(\theta -45°\right)$is 1.

The period of $y=\cos \left(\theta -45°\right)$is $2\pi$.

The phase shift of $y=\cos \left(\theta -45°\right)$is $45°$.

Step-by-step solution

Step 1. Write down the given information.

The given function is $y=\cos \left(\theta -45°\right)$.

Step 2. Concept used.

A function of the form:

$y=a\sin b\left(\theta -h\right)+k,y=a\cos b\left(\theta -h\right)+k\text{and}y=a\tan b\left(\theta -h\right)+k$ has vertical shift $\left(k\right)$. And, a period of $\frac{360°}{\left|b\right|}\text{or}\frac{2\pi }{\left|b\right|}$ for sine, cosecant, secant and cosine functions and a period of $\frac{180°}{\left|b\right|}\text{or}\frac{\pi }{\left|b\right|}$ for tangent and cotangent function. The phase shift for the functions is $\left(h\right)$.

The amplitude of secant, cosecant, tangent and cotangent functions is not defined.

The equation for the midline is written as, $y=k$. A midline is a new reference line when the parent graph is stretched vertically up or down and then the graph oscillates about new reference line called the midline.

Step 3. Calculation.

With the help of concept stated above, the amplitude, period and phase shift of the function is evaluated as:

The amplitude of $y=\cos \left(\theta -45°\right)$is $\left|1\right|=1$.

The period of $y=\cos \left(\theta -45°\right)$ is $\frac{2\pi }{\left|1\right|}=2\pi$.

The phase shift of $y=\cos \left(\theta -45°\right)$ is $45°$.

Step 4. Sketch the graph of the function.

The graph of the function $y=\cos \left(\theta -45°\right)$ is shown below.

Step 5. Conclusion.

The amplitude of $y=\cos \left(\theta -45°\right)$is $1$.

The period of $y=\cos \left(\theta -45°\right)$ is $2\pi$.

The phase shift of $y=\cos \left(\theta -45°\right)$ is $45°$.