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Trig question.

  1. Oct 1, 2006 #1
    I don't really understand a step in my textbook, and I think it is better that I understand why they are doing it this way instead of just brushing it off and it coming back to haunt me later!

    The question is : Solve for [tex]\Theta, 0\leq\Theta\leq2\pi[/tex]


    I get the right answer, but their last step confuses me....

    This is how they show their work :



    [tex]\Theta=\frac{\pi}{4}[/tex] I don't fully grasp this step...

    I understand that tan is 1 at [tex]\frac{\pi}{4}[/tex] since sin and cosine are both [tex]\frac{\sqrt{3}}{2}[/tex] at that point. But how can you call tan pi/2 when it also is 1 at [tex]\frac{5\pi}{4}[/tex]... Can someone explain what they are doing? They don't even give [tex]\frac{5\pi}{4}[/tex] as an answer in their equation, they just state it later in a sentence after the question.

    Last edited: Oct 1, 2006
  2. jcsd
  3. Oct 1, 2006 #2
    Nevermind I understand now, fairly simple. Just convert the equation to [tex]\Theta=tan^{-1}(1)[/tex]


    This trigonometry stuff is slowly eating my brain, lol. ahh!
  4. Oct 1, 2006 #3
    I guess I don't understand. [itex]\frac{5\pi}{4}[/itex] and [itex]\frac{\pi}{4}[/itex] both work, they are both solutions to the equation [itex]2tan\theta -2 = 0[/itex]. Reducing the equation to [itex]\theta=tan^{-1}(1)[/itex] isn't the answer since you have to pick which part of tangent you want to invert. Putting [itex]tan^{-1}(1)[/itex] into your calculator will give you [itex]\frac{\pi}{4}[/itex] but this is not the only answer. It would be if your inequality read something like [itex]0\leq \theta \leq \frac{\pi}{2}[/itex] but it doesn't. You're looking for all solutions between zero and [itex]2\pi[/itex]. I'm not sure why your book doesn't include both solutions though, that's odd. But you were right to be bothered by this.
  5. Oct 1, 2006 #4
    hmm....[tex]\frac{\pi}{4}[/tex] is your alpha.after getting your alpha,you are suppose to use it in the general solution and for tan,it is n[tex]{pi}[/tex]+[tex]\frac{\pi}{4}[/tex] where n is all real number.since the question ask for [tex]\Theta, 0\leq\Theta\leq2\pi[/tex], the only 2 are [tex]\frac{5\pi}{4}[/tex] and [tex]\frac{\pi}{4}[/tex] coz your n can only take 0 and 1.get it?
    Last edited: Oct 1, 2006
  6. Oct 1, 2006 #5
    I guess that makes sense :approve: I just thought that they should show that in their algebraic answer, but what you are saying makes sense and seems to coincide with their instructions. Thanks
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