# Solving sin(ax-b)

1. Aug 24, 2015

### adjacent

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

### adjacent

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

### adjacent

Make it 2 :) This method is easy
Looks like I somehow did my simultaneous equation wrong.

I got the answer thanks SteamKing and mfb :)

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