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Trig and Unit Circles

  1. Feb 19, 2006 #1
    is it possible to solve cos pi(t) + sin pi(t) = 0 for determining the value of pi(t)??
    pls help
  2. jcsd
  3. Feb 19, 2006 #2
    Yes. Think of it this way: you want values of [itex]\theta = pi(t)[/itex] on the unit circle such that [itex]\cos(\theta)=-\sin(\theta)[/itex].
  4. Feb 19, 2006 #3
    ya,exactly what i wanted to find but i have thought of this way oso but from here i can't get to contiue already
  5. Feb 19, 2006 #4


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    Why not draw the unit circle and try a few points, see if that gives you any idea.
  6. Feb 19, 2006 #5
    I think you just need to know your unit circle and values for the sin and cos. The only other way I can think of is trying to solve for pi(t) via inverse trig functions but I'm not getting any simplifications from Mathematica using that tactic.
  7. Feb 19, 2006 #6


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    If you want you could solve it algebraically by squaring both sides of cos x = -sin x and then expressing cos^2 x in terms of sin^2 x, then rearranging terms and using an inverse function, but that's more complicated than you need and it might make you miss one of the solutions if you're not careful. You can solve this one just by looking at the unit circle.
  8. Feb 20, 2006 #7


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    Or you may try to do the following:
    If [tex]\cos (\pi (t)) = 0[/tex] then [tex]\sin (\pi (t)) = \pm 1[/tex]
    So [tex]\cos (\pi (t)) + \sin (\pi (t)) = 0 \pm 1 = \pm 1 \neq 0[/tex]
    That means if [tex]\cos (\pi (t)) = 0[/tex] then the LHS is not 0, and hence it does not satify the equation.
    So it's true that [tex]\cos (\pi (t)) \neq 0[/tex].
    Divide both sides of the equation by [tex]\cos (\pi (t))[/tex] to obtain:
    [tex]1 + \tan (\pi (t)) = 0[/tex]
    Now, can you go from here? :)
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