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Solving trig function (possible error in solutions)

  • #1
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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..
 

Answers and Replies

  • #2
I like Serena
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What is tan(-3pi/4)?
 
  • #3
dynamicsolo
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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.
 
  • #4
SammyS
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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..
Are you sure the problem isn't, [itex]\sin^2(t)=\cos^2(t)\,?[/itex]
 

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