# Homework Help: Solving sin(ax-b)

1. Aug 24, 2015

1. The problem statement, all variables and given/known data

The figure shows part of a curve with the equation $y=sin(ax-b)$ where $a>0$ and $0<b<\pi$. The curve cuts the x-axis at the points P, Q and R as shown.
Given that the coordinates of P, Q and R are
$\Big(\frac{\pi}{10},0\Big),\Big(\frac{3\pi}{5},0\Big)$ and $\Big(\frac{11\pi}{10},0\Big)$ respectively. Find the values of $a$ and $b$

2. Relevant equations

3. The attempt at a solution
I know how to solve the this if $ax-b$ was simply $x$.... $sin^{-1}(0)$ gives 0. So at the interval $0\leq x \leq 2\pi$ ,$x$ will be $\pi-0$(Which is pi)$, 0,2\pi$
But for this, I did:
$\text{let }\alpha = ax-b$
$sin^{-1}(0)=0$
So $\alpha= 0,\pi,2\pi$
Looking at the diagram, first point is $\Big(\frac{\pi}{10},0\Big)$ and the value of $x$ is $\frac{\pi}{10}$. Also the first solution for $\alpha$ is 0. Therefore:
$\alpha=0$
$a\frac{\pi}{10}-b=0$
$a\frac{3\pi}{5}-b=\pi$
Solving this simultaneously gives $a=\frac{10}{3}$ which is wrong. The actual answer is $a=2,b=\frac{\pi}{5}$ so my method is wrong somehow.
So what should I do?

Last edited: Aug 24, 2015
2. Aug 24, 2015

### SteamKing

Staff Emeritus
Hints:

1. The difference between R and P should equal the period of the sine function.
2. The quantity "b" is just a phase angle, which has no effect on the period of the function.

3. Aug 24, 2015

But what has this to do with the period of the sine function? Period of $sin(ax-b)$ is $\pi$ . What next?

4. Aug 24, 2015

### SteamKing

Staff Emeritus
The period of sin (x) is 2π. What must you do to the argument x to change the period of the sine function from 2π to just π?

5. Aug 24, 2015

### Staff: Mentor

Looks fine up to this point.
How did you get that answer?

6. Aug 24, 2015