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Trig identies

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Sin(2α)=-8/17
    Using either double angle or half angle identities, sin,cos,tan,csc,sec and cot of alpha.



    2. Relevant equations



    3. The attempt at a solution

    I tried and tried BUT i am lost because in order to find cos of alpha using identies, you still have to have cos2a or cos^2a. I have not been able to find an identy that just uses sin(2a) and spits out some sort of cos. I tried using cos2a=1-2sin^2alpha but did not end up with anything close to what the book says is the answer. Funny thing is, what ever algerbra i try on the different identities, i still get the same answer that i got previously. the book says: 4√17/17, √17/17, -4, √17/4, -√17, and -1/4 but doesnt tell me if that is cos or what function it is. haha. i dont have the solutions manual either because that was an extra 50$ i couldnt afford.

    any help would be nice. pretty pics IF possible but i understand if not. thanks
     
  2. jcsd
  3. Oct 15, 2011 #2

    SammyS

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    It's not very clear.

    What are you trying to find?

    If sin(θ) = -8/17, what is cos(θ) ?
     
  4. Oct 15, 2011 #3

    i wrote it exactly as the book stated it.Haha. i agree its unclear. it gave me sin(2a) and wants me to find sin(a) cos(a) csc(a) sec(a) tan(a) cot(a)

    i have to use either a double angle identity or a half angle identy

    i cant figure out which one to use because there is not an identy that calls for just sin. they all ask for at least cos and or sin.

    thanks
     
  5. Oct 15, 2011 #4

    VietDao29

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    I think he's trying to derive the value of sin(θ), cos(θ), tan(θ), and cot(θ) using sin(2θ) = -8/17.

    Yes, this is true, if you have cos(2α), then finding sin(α), cos(α), tan(α), and cot(α) should be easy as a piece of cake, right?

    So the problem remains is, from sin(2α), how can you find cos(2α)? Hint: You should consider using Pythagorean Identity:

    [tex]\sin ^ 2 (\gamma) + \cos^2(\gamma) = 1[/tex]

    Can you go from here? :)
     
  6. Oct 15, 2011 #5
    thanks

    yeah cos(2a) is a piece of cake. alot of writting BUT i know where to go with it. however, i spent 2 hours on this opne last night and got no where. I kept getting a different answer BUT the same answer everytime. so my results were consistant.

    i have looked at that one for sin^2a=1-cos^2a but i dont end up with the same answer as they do. BUT my algerbra indeed is not that great. i still struggle with fractions. haha. if you can or would like to. do you get any of the previous answers i posted in my original post when using the Pythagorean therom?

    thanks
     
  7. Oct 15, 2011 #6

    VietDao29

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    Yes, I do.

    Can you show me your work, so that I can check it for you?
     
  8. Oct 15, 2011 #7

    dang... ok let me try it again and ill show work. thanks. give me a few please
     
  9. Oct 15, 2011 #8

    SammyS

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    Well, writing that you need to find sin(a) cos(a) csc(a) sec(a) tan(a) cot(a), makes it much clearer than saying "Using either double angle or half angle identities, sin,cos,tan,csc,sec and cot of alpha."

    (I see that you did end that with "alpha", at the end of that run-on string.)

    If cos(2α) is a piece of cake, (except for the sign), then use cos(2x) = 2cos2(x) - 1 = 1 - 2sin2(x), to find cos(α) and sin(α)
     
  10. Oct 15, 2011 #9
    i agree that it was sloppy. sorry.

    i have tried both of those ways and i get 13/17 or 255/289.

    if sin(2a)= -8/17 then sin(a) should be 4/17, if in Q2, since sin is half of 2 alpha.

    cos^2a=1-(4/17)^2 so that would be 289/289-16/289 then root that which is not anything the book gets.

    using the other one.

    i get this:

    1-2(4/17)2

    i get 1-2(16/289)

    then that goes to 1-32/578

    which goes to 578/578-32/578

    so im not seeing how this is getting to where i need to be.
     
  11. Oct 15, 2011 #10

    eumyang

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    This is not correct. You are confusing sin(2α) with 2sin(α), which are not necessarily equal.

    If sin2θ + cos2θ = 1, then surely
    sin2(2α) + cos2(2α) = 1.
    Substitute:
    (-8/17)2 + cos2(2α) = 1
    ... and solve for cos(2α).

    Then use the identity
    cos(2α) = 2cos2(α) - 1
    to find cos(α). The rest should follow.
     
  12. Oct 15, 2011 #11

    SammyS

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    I think you said you got cos(2α). If so, what did you get for it.

    cos(2α) = ___?___
     
  13. Oct 15, 2011 #12
    but isnt sin(a) half of sin(2a)? because if you go sin(2a/2) that should canle the 2's out leaving only sin(a):

    so sin(2a)=-8/17

    then sin(2a/2)=8/17/2 which should yield sin(a)=8/34 or 4/17

    how is dividing -8/17 giving me 2sin(a)? i dont see it at all.

    thanks
     
  14. Oct 15, 2011 #13
    i did BUt it is not what the others or the book has gotten.
     
  15. Oct 15, 2011 #14

    SammyS

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    The problem doesn't ask for cos(2α) .
     
  16. Oct 15, 2011 #15

    eumyang

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    No, it isn't!
    You can't divide both sides by 2 like that.

    There is another identity involving sin(2α):
    sin(2α) = 2sinsin(α)cos(α)
     
  17. Oct 15, 2011 #16
    it does because in order to get to cos(a), i need to know cos(2a) i have been trying to find cos(2a) so i can use the other identity. BUT i cant even get past the whole sin(2a) thing for some reason.
     
  18. Oct 15, 2011 #17

    SammyS

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    sin2(x) + cos2(x) = 1

    Therefore, cos(x) = √(1 - sin2(x) )
     
  19. Oct 15, 2011 #18
    see now im confused. how can 1a not be half of 2a? i dont see how dividing sin(2a) in half is wrong or not the same as sin(1a). Is it because the 2 would also be under teh sin as well since everything is attached to that?


    i cant use that identity because i dont know cos(a) yet. lol
     
  20. Oct 15, 2011 #19

    eumyang

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    sin(2x) ≠ 2sin(x)
    Try putting in some angles and see!
    sin(45°) ≈ 0.7071
    sin(90°) = 1
    2 * 0.7071 ≠ 1!

    SammyS, VietDao29, and I have been nudging you a certain direction. Look at my previous post #10:
    ... and please try it out.
     
  21. Oct 15, 2011 #20
    i have done that one and it doesnt work. sorry for not understanding at all but it makes no sense to me. i have tried substituting it into the normal pythagerian formula nad still get the same answer as i always get

    √1-(-8/17)2

    goes to:

    √1-64/289
    √289/289-64/289
    √15/17

    which is not on there
     
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