# Find the value of $A$, $n$ and $b$ if $y=A\sin(nt)+b$ has range $[2,8]$ and period $\frac{2\pi}{3}$.

A function with rule $y=A\sin(nt)+b$ has range $[2,8]$ and period $\frac{2\pi}{3}$. Find the value of $A$, $n$ and $b$.

According to the teacher tip

• Do Dilations before translations

But found translations first and I got $n$ , it is right?

$\frac{2\pi}{3}n$=$2\pi$
$n=3$

But I don't know how to find $A$ and $b$.
Many thanks.

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You know that $-1 \leq \sin\,x \leq 1$, no? What happens if you multiply by $A$ all sides of that inequality? –  Ｊ. Ｍ. May 13 '12 at 12:19
sorry, no. I'm confuse. thx –  Sb Sangpi May 13 '12 at 12:25
You don't need to think about translations since that's already given when it says that it's of the form $y=A\sin(nt)+b$. –  Michael Hardy May 13 '12 at 17:38

## 2 Answers

We know that $\sin(x)$ takes values in the range $[-1, 1]$. When $\sin(nt)$ attains its maximum of $1$, we want $y$ to be $8$. Plug the values into the equation to get:

$$8 = A + b$$

Similarly, when $\sin(nt)$ is $-1$, we want $y$ to be $2$:

$$2 = -A + b$$

Now, you have a system of 2 equations with 2 variables. Solve to get $A$ and $b$.

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We know that:

$$-1 \leq \sin (3t) \leq 1$$

Multiplying both sides by $A$:

$$-A \leq A\sin (3t) \leq +A$$

Then by adding $b$ to both sides, we have:

$$-A+b \leq A\sin (3t)+b \leq A+b \tag{1}$$

Comparing $(1)$ to what you have from the question that $2 \leq y \leq 8$, you can easily find the values of $A$ and $b$ from the system of equations:

$$\left\{ \begin{array}{ll} -A+b=2 & \\ A+b=8 & \end{array} \right.$$

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