Solve Sin(ax-b) Homework: Find Values of a & b

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Homework Statement


ddd.png

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##

Homework Equations

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?
 
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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.
 
But what has this to do with the period of the sine function? Period of ##sin(ax-b)## is ##\pi## . What next?
 
adjacent said:
But what has this to do with the period of the sine function? Period of ##sin(ax-b)## is ##\pi## . What next?
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?
 
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adjacent said:
##a\frac{\pi}{10}-b=0##
##a\frac{3\pi}{5}-b=\pi##
Looks fine up to this point.
Solving this simultaneously gives ##a=\frac{10}{3}## which is wrong
How did you get that answer?
 
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SteamKing said:
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?
Make it 2 :) This method is easy :biggrin:
mfb said:
Looks fine up to this point.
How did you get that answer?
Looks like I somehow did my simultaneous equation wrong.

I got the answer thanks SteamKing and mfb :)
 

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