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

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Homework Help Overview

The problem involves analyzing the sine function represented by the equation ##y=sin(ax-b)##, where ##a>0## and ##0

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the sine function's period and the given points of intersection. There is an exploration of how to set up equations based on the intersections and the implications of the phase shift represented by ##b##.

Discussion Status

Some participants have offered hints regarding the period of the sine function and its relationship to the values of ##a## and ##b##. There is an ongoing examination of the initial attempts to solve the equations, with some questioning the correctness of the derived values.

Contextual Notes

Participants note that the period of the sine function is relevant to the problem, and there is a focus on the implications of the phase angle ##b##. The original poster expresses uncertainty about their method and results, indicating a need for further clarification.

adjacent
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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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