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Simple trig functions

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data
    (In radians I assume) Using the exact values of sin, cos, tan of 1/3 pi and 1/6 pi, and the symmetry of the graphs of sin, cos and tan, find the exact values of sin(-1/6pi), cos (5/3pi) and tan(4/3pi).


    2. Relevant equations
    cos(x)=sin(x+pi/2)
    sin(x) = sin(x+2pi)
    sin^2(x)+cos^2(x)=1


    3. The attempt at a solution

    I have a little difficulty understanding what is required here. I have a list of trig functions but these don't seem to be here. I imagine the answer is some square root with a coefficient, but I'd like to understand it better. Can someone point me in the right direction without just giving me the answer? Thanks
     
  2. jcsd
  3. Jan 6, 2010 #2

    rock.freak667

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    Try the numbers as combination of π, for example to find tan(5π/4)=tan(π+π/4). Like that.
     
  4. Jan 6, 2010 #3
  5. Jan 6, 2010 #4

    vela

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    The idea is to look at the graphs of functions to figure out properties that would be useful in solving the problems. For example, if you graph [tex]\sin x[/tex], you'll see that it's an odd function; in other words, it has the property [tex]\sin(x) = -\sin(-x)[/tex]. If [tex]x=-\pi/6[/tex], you can say [tex]\sin(-\pi/6) = -\sin[-(-\pi/6)] = -\sin(\pi/6)[/tex].
     
  6. Jan 6, 2010 #5
    Thanks guys. I guess I'm being slow, but I've had a few tries now and can't equate, e.g. sin(-pi/6) with either sin, cos, or tan of 1/3 or 1/6 pi.
    For example:

    sin(-pi/6) = -sin(pi/6)
    fine.
    cos(pi/6 + pi/2) = -sin(pi/6)
    so
    cos(4/6 pi) = sin(-pi/6)

    but what I really want to equate the LHS to is
    -cos(1/3 pi).
    I know I'm being really slow, but I can't seem to get there.
     
  7. Jan 6, 2010 #6
    The graph does help a lot.
    I can see that sin(-1/6 pi) is equal to cos exactly half pi earlier in the graph. So is there an identity that states sin(x - pi/2)=cos(x)?
    If so, I've cracked it.
     
  8. Jan 6, 2010 #7
    Sorta but not quite.

    Notice if you put x = 0, you'd get
    sin (-pi/2) = cos 0 = 1 which is wrong. The right identity.

    It'd be
    [tex]
    \sin (\frac{\pi}{2} - x ) = \cos x
    [/tex]
     
  9. Jan 6, 2010 #8

    rock.freak667

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    should be sin(x-π/2)= -cos(x)
     
  10. Jan 6, 2010 #9

    vela

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    Actually, I think Tom discovered the identity

    [tex]sin(x+\frac{\pi}{2}) = cos x[/tex]

    which just says that the graph of cosine is the sine shifted by [tex]\pi/2[/tex]. It's easy to screw up the sign of the phase shift because, to me, it seems like it should be the opposite sign intuitively. I always have to double-check to make sure I get it right.
     
  11. Jan 6, 2010 #10

    vela

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    You probably can't because that's actually kind of hard to do. :)

    The problem wants you to figure out what [tex]\cos(5\pi/3)[/tex] equals in terms of [tex]\cos(\pi/6)[/tex] or [tex]\cos(\pi/3)[/tex]. Same with [tex]\tan(4\pi/3)[/tex] and [tex]\tan(\pi/6)[/tex] or [tex]\tan(\pi/3)[/tex].
     
  12. Jan 6, 2010 #11
    Thanks all.
    Well I think I'm getting there.
    I think for the first one I can just say

    sin(pi/6)=0.5
    sin (-x) = -sin(x), therefore
    sin(-pi/6)=-sin(pi/6)=-0.5

    Simple.

    For the second one, I'm guessing I have to use:
    cos(pi/3)=0.5
    and somehow show that it is equal to:
    cos(5pi/3).

    I can see the graph very clearly, and understand why this is the case. But I'm not sure how I can show it. I know that as the angle goes from 0 to pi, the cosine of the angle goes from 1 to -1, and then as the angle goes from pi to 2pi, the cosine goes back round from -1 to 1. So of course, the cosine at 1/3 of the way round to pi will be the same as it is at 2pi - 1/3pi. But how to show that mathematically?

    Thanks again, and sorry for being slow-witted.
     
  13. Jan 6, 2010 #12

    rock.freak667

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    5π/3 = 2π- π/3

    Use this.
     
  14. Jan 6, 2010 #13

    vela

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    So essentially what you've figured out is that [tex]\cos x = \cos(2\pi - x)[/tex]. This another one of the many trig identities. You can just say

    [tex]\cos 5\pi/3 = \cos(2\pi-\pi/3) = \cos \pi/3 = ...[/tex]
     
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