Trigonometry Engineering Waves proble

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Homework Help Overview

The discussion revolves around a trigonometry problem related to engineering waves, specifically transforming the equation y = sin t - cos t into the form A cos(ωt + α). The original poster seeks clarification on the derivation of the angle α, particularly why it is expressed as 5π/4.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the tangent function and its values in different quadrants, particularly focusing on the third quadrant where the angle α is derived. Questions arise about the use of the cast rule and how different quadrants correspond to specific values of π.

Discussion Status

Participants are actively engaging with the problem, providing insights into the tangent function and its behavior in various quadrants. Some guidance has been offered regarding the relationship between angles and their respective quadrants, though a complete consensus on the understanding of these concepts has not been reached.

Contextual Notes

There is a noted uncertainty regarding the interpretation of quadrant values in terms of π, and participants express a desire to clarify these concepts further. The original poster references a past paper solution that lacks detailed explanations, contributing to the confusion.

andycampbell1
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Homework Statement


Hi

I am doing a maths past paper and I would like some help to understand where part of the solution came from. The part I do not understand is the bit at the end which says tan[tex]\alpha[/tex] = 5[tex]\pi[/tex]/4. It would be great if someone could help me understand where this came from.

The question is

y = sin t - cos t in the form A cos([tex]\omega[/tex]t+[tex]\alpha[/tex])


Homework Equations





The Attempt at a Solution



The solution is

y=sin t -cos t = A cos([tex]\omega[/tex]t+[tex]\alpha[/tex])
= A cos t cos [tex]\alpha[/tex] - A sin t sin [tex]\alpha[/tex]

A cos [tex]\alpha[/tex] = -1
- A sin [tex]\alpha[/tex] = 1

Squaring the above equations and adding side by side implies that A = [tex]\sqrt{2}[/tex]

Dividing the above equations side by side gives

tan [tex]\alpha[/tex]= 1

tan is positive in the first and third quadrant. We choose the third quadrant since from the above equation sin and cos are negative.

Hence [tex]\alpha[/tex] = [tex]\pi[/tex]tan-1 = 5[tex]\pi[/tex]/4

y= sin t - cos t = [tex]\sqrt{2}[/tex]cos(t+ 5[tex]\pi[/tex]/4)

What I do not understand is where did 5[tex]\pi[/tex] come from? I have checked my notes and it does not mention anything about where pi comes from.
 
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hi andy! :smile:

(have an alpha: α and a pi: π and an omega: ω and a square-root: √ :wink:)

tan(π/4) = tan(45°) = 1

tan(π + x) = tanx, so tan(π + π/4) = 1 also, ie tan(5π/4) = 1 :smile:
 
Thanks. If I was using the cast rule does each quadrant have its own value of pi? I get the first part but I don't get the second line. Sorry ha I'm just hopeless at trigonometry.
 
andycampbell1 said:
Thanks. If I was using the cast rule does each quadrant have its own value of pi? I get the first part but I don't get the second line. Sorry ha I'm just hopeless at trigonometry.

You just said you have chosen the third quadrant therefore pi + pi/4 is the value you are looking for because it's in the third quadrant.
 
I know how to work out what quadrant to put tan in but. I do not know what each quadrant represents in terms of pi. Like u said pi in the 3rd quadrant = pi
+ pi/4. What would the other quadrants wqual
 
I know how to work out what quadrant to put tan in but. I do not know what each quadrant represents in terms of pi. Like u said pi in the 3rd quadrant = pi
+ pi/4. What would the other quadrants equal? The solution I posted came with the past paper I am just trying to make sense of it.
 
Each quadrant has 90o (or [tex]\pi/2[/tex] radians). The quadrant ranges are:

First quadrant ranges from 0 to [tex]\pi/2[/tex]
The second from [tex]\pi/2[/tex] to [tex]\pi[/tex]
The third from [tex]\pi[/tex] to [tex]3\pi/2[/tex]
The fourth from [tex]3\pi/2[/tex] to [tex]2\pi[/tex]
 
Thanks, from the examples I have the quadrant values make them make sense to me now, thanks for your help.
 

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