Solving trig function (possible error in solutions)

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Homework Help Overview

The discussion revolves around solving a trigonometric equation involving sine and cosine functions. The original poster expresses confusion regarding the solutions provided in their textbook, particularly concerning the inclusion of both positive and negative values for the angles.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the equation sin(2t) = cos(2t) and the resulting tangent function. There is a discussion about the nature of solutions in trigonometric equations, particularly regarding the periodicity and the inclusion of negative angles.

Discussion Status

Some participants suggest that the original poster consider the periodic nature of the tangent function, indicating that multiple solutions exist beyond the principal values. There is also a question raised about the possibility of a different problem statement involving sine squared and cosine squared.

Contextual Notes

Participants note that the textbook's solutions include both positive and negative angles, which may not align with the original poster's interpretation of the problem. The discussion hints at the potential for misunderstanding the problem setup or the nature of the solutions.

sandy.bridge
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Homework Statement


Alright, so I am going through the solutions to one problem that has me stumped, and initial instinct tells me it is wrong, however, I figured I would get clarification.


The Attempt at a Solution


[tex]sin(2t)=cos(2t)[/tex]
[tex]tan(2t)=1[/tex]
[tex]2t=\pi{/}4, 5\pi{/}4[/tex]

However, the book states both positive and negative values, rather than just positive. It's either not correct, or not clicking. tan(-π/4)=-1, not 1..
 
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What is tan(-3pi/4)?
 
Since your text is suggesting that there are positive and negative solutions, you are not restricted to the "principal circle" , [itex]0 ≤ t < 2{\pi}[/itex] . So you can "go around the circle" as many times as you like:

[tex]2t = \frac{\pi}{4} + k \cdot 2{\pi} , \frac{5\pi}{4} + k \cdot 2{\pi} ,[/tex]

with k being any integer.
 
sandy.bridge said:

Homework Statement


Alright, so I am going through the solutions to one problem that has me stumped, and initial instinct tells me it is wrong, however, I figured I would get clarification.


The Attempt at a Solution


[tex]sin(2t)=cos(2t)[/tex]
[tex]tan(2t)=1[/tex]
[tex]2t=\pi{/}4, 5\pi{/}4[/tex]

However, the book states both positive and negative values, rather than just positive. It's either not correct, or not clicking. tan(-π/4)=-1, not 1..

Are you sure the problem isn't, [itex]\sin^2(t)=\cos^2(t)\,?[/itex]
 

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