We are given the graph of a sine function, and want to find its period and write its equation. Let's do these things one at a time.

Finding the Period

Recall that the period of a function is the horizontal length of one cycle. Let's pay close attention to the graph to note the period and the maximum and minimum values.

According to the graph, one cycle takes $π$ radians to complete. Therefore, the period of the function is $π .$ We also see that the maximum value of the graph is $\frac{5}{2},$ and that the minimum value is $- \frac{5}{2} .$ Therefore, the amplitude is $\frac{5}{2} .$

Writing the Equation

Let's recall the general form of a sine function.

y = a sin b θ

Here,

∣ a ∣

is the amplitude and

b

is a nonzero number such that

\frac{2 π}{b}

is the period of the function. Since we already know that the period is

π,

we can write an equation in terms of

b .

\frac{2 π}{b} = π

Let's solve this equation for

b .

\frac{2 π}{b} = π

▼

Solve for

b

MultEqn

$LHS \cdot b = RHS \cdot b$

2 π = b π

DivEqn

$LHS / π = RHS / π$

2 = b

RearrangeEqn

Rearrange equation

b = 2

We found that

b = 2 .

y = a sin b θ \Rightarrow y = a sin 2 θ

We previously found that the amplitude is

\frac{5}{2} .

Looking back at the graph, we can see that the maximum value occurs before the minimum in the interval

0

to

2 π .

Therefore,

a > 0

and we can say that

a = \frac{5}{2} .

y = a sin b θ \Rightarrow y = \frac{5}{2} sin 2 θ

Exercise

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