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Pythagorean Identities - finding sin and cos ?

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    For tan(theta) = /2/3 (theta in quadrant 2) find sin(pheta) and cos(pheta) using the pythagorean identities.



    2. Relevant equations



    3. The attempt at a solution

    Which 2 pythagorean identities would I use?

    My trouble is with finding sin and cos from tan.
     
  2. jcsd
  3. Jun 26, 2011 #2

    Dick

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    You probably want to use tan(theta)=sin(theta)/cos(theta) and sin(theta)^2+cos(theta)^2=1. Now try it.
     
  4. Jun 26, 2011 #3
    So, I get this one here: sin(theta)^2+cos(theta)^2=1 - worked out :)

    But my other 2 options, according to my text's formula sheet, I can choose from the following formulas?

    1+tan^2(theta) = sec^2(theta)

    1 + cot^2(theta) = csc^2(theta)

    ?
     
  5. Jun 26, 2011 #4

    Dick

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    Since your problem involves tan(theta), I'd try working with the first one. sec(theta)=1/cos(theta), yes?
     
  6. Jun 26, 2011 #5
    Dick, there is something I am not getting.

    My formula sheet has the following 2 formulas to use, and the formula you are telling me is not one of them?

    1+tan^2(theta) = sec^2(theta)

    1 + cot^2(theta) = csc^2(theta)

    Thanks so much for your help.
     
  7. Jun 26, 2011 #6

    Dick

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    sec(theta)=1/cos(theta) may not be in your list of pythogorean identities along with tan(theta)=sin(theta)/cos(theta), but I think you can use them anyway. They are sort of the definition of sec(theta) and tan(theta).
     
  8. Jun 26, 2011 #7

    eumyang

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    I guess you could start by using the
    1 + tan2 θ = sec2 θ
    identity to find sec θ, and by extension, cos θ. Then use another Pythagorean identity to find sin θ.
     
  9. Jun 27, 2011 #8

    SteamKing

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    Take the Pythagorean Identity. Divide thru using the sin^2 term. Simplify.
    Take the Pythagorean Identity. Divide thru using the cos^2 term. Simplify.
    You should derive the two formulas on your formula sheet. They are both restatements of this most important trig identity.
     
  10. Jun 27, 2011 #9

    HallsofIvy

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    The 'basic' Pythagorean identity is just the Pythagorean theorem- if a right triangle has legs of length a and b and hypotenuse of length c, then [itex]c^2= a^2+ b^2[/itex].


    If tan(theta)= 2/3, then the triangle is similar to a triangle with "opposite side" of length 2 and "near side" of length 3. By the Pythagorean theorem, the hypotenuse, c, is given by [itex]c^2= 2^2+ 3^2= 4+ 9= 13[/itex] so the length of the hypotenuse is [itex]\sqrt{13}[/itex]. Now, knowing that you have a right triangle with "opposite side" of length 2, "near side" of length 3, and "hypotenuse" of length [itex]\sqrt{13}[/itex] you can immediately calculate any of the trigonometric functions of this angle.
     
  11. Jun 27, 2011 #10

    tiny-tim

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    hi nukeman! :smile:

    (have a theta: θ and try using the X2 icon just above the Reply box :wink:)
    each of the three Pythagorean identities only relates two functions at a time …

    one relates tan and sec, one relates cot and cosec, and one relates cos and sin …

    to start from tan and get cos, you need to go from tan to sec (as eumyang :smile: says), then use sec = 1/cos

    to start from tan and get sin, you either need to do al that, and then go from cos to sin, or you can start again, use cot = 1/tan, then go from cot to cosec, then use sin = 1/cosec :wink:
     
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