Q3. - Algebra 2

Question

Write the equation of a trigonometric function with a phase shift of $-45°$.

Step-by-Step Solution

Answer

The equation of the trigonometric function with a phase shift of $-45°$ is $y=\sin \left(\theta +45°\right)+2$.

Step 1. Write down the given information.

The given statement is to define the concept of midline of trigonometric graph. 

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 phase shift $\left(h\right)$. And if $h>0$ the phase shift is to the right and if $h<0$ then the phase shift is to the left.

Step 3. Write the equation of the trigonometric function with a given phase shift.

With the help of concept stated above, the equation of trigonometric function with a phase shift of $-45°$ is written as:

 $\begin{array}{c}y=a\sin b\left(\theta -h\right)+k\\ y=\left(1\right)\sin \left[1\left(\theta -\left(-45°\right)\right)\right]+\left(2\right)\mathrm{....}\left(\begin{array}{l}\text{Plugging for }h=-45°,\\ a=1,b=1\text{ and }k=2\end{array}\right)\\ y=\sin \left(\theta +45°\right)+2\end{array}$

Step 4. Conclusion.

The equation of the trigonometric function with a phase shift of $-45°$ is $y=\sin \left(\theta +45°\right)+2$.