Q33. - Algebra 2

Question

Draw a graph of sine function with an amplitude of $\frac{3}{5}$ and a period of $90°$. Then write an equation for the sine function.

Step-by-Step Solution

Answer

The equation of the sine function is $y=\frac{3}{5}\sin \left(4x\right)$.

The amplitude of $y=\frac{3}{5}\sin \left(4x\right)$ is $\frac{3}{5}$.

The period of $y=\frac{3}{5}\sin \left(4x\right)$ is $90°$.

Step 1. Write down the given information.

The given sine function has an amplitude of $\frac{3}{5}$ and a period of $90°$.

Step 2. Concept used.

A function of the form $y=a\sin \left(bx\right) \text{and }y=a\cos \left(bx\right)$ has amplitude of $\left|a\right|$ and a period of $\frac{360°}{\left|b\right|} \text{or }\frac{2\pi }{\left|b\right|}$.

Step 3. Write the equation of sine function of given amplitude and period.

Since the given sine function has an amplitude of $\frac{3}{5}$ and a period of $90°$. Therefore, plugging $\left|a\right|=\frac{3}{5}$ in sine function $y=a\sin \left(bx\right)$ and equating period of sine function $\frac{360°}{\left|b\right|}=90°$ to get $b=4$.

Now, plugging $\left|a\right|=\frac{3}{5}$ and $b=4$ in $y=a\sin \left(bx\right)$. Therefore, the equation of sine function of given amplitude and period is written as $y=\frac{3}{5}\sin \left(4x\right)$.

Step 4. Sketch the graph for the obtained function.

The graph for the function $y=\frac{3}{5}\sin \left(4x\right)$ of amplitude $\left|a\right|=\frac{3}{5}$ and period $90°$ is shown below.

Step 5. Interpretation from the graph.

The equation of the sine function is $y=\frac{3}{5}\sin \left(4x\right)$.

The amplitude of $y=\frac{3}{5}\sin \left(4x\right)$ is $\frac{3}{5}$.

The period of $y=\frac{3}{5}\sin \left(4x\right)$ is $90°$.