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Trig Identity problem

  1. Dec 1, 2006 #1
    Hey ive got an assignment on trig identities and cant figure this one out.

    Prove the Identity:


    I got to

    tanx+cotx= 1/2 sin2x=1/4sinx2cosx

    but when i get to the point where i have numbers in front of the sinx or cosx i dont know what to do.

    Thanks for any help
  2. jcsd
  3. Dec 1, 2006 #2


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    try multiplying tan(x) + cot(x) by tan (x) and proceeding from there. I'll start you off,

    [tex] tan(x) + cot(x) = \frac{sec^2(x)} {tan(x)} [/tex]
  4. Dec 1, 2006 #3
    Write the LHS in terms of [tex] \sin x [/tex] and [tex] \cos x [/tex]. You should get:

    [tex] \frac{1}{\sin x \cos x} = \frac{1}{\frac{1}{2}\sin 2x} = \frac{2}{\sin 2x} = 2\csc 2x [/tex]
    Last edited: Dec 1, 2006
  5. Dec 2, 2006 #4
    im sorry but i still have no idea how to do this question
  6. Dec 2, 2006 #5
    have you tried changing everything to sin and cos ? doing that you will be able to get the right side to equal the left.

    if you're still stuck just post what you've tried to do and someone could point out where you went wrong or give you some advice on what the next step would be
  7. Dec 2, 2006 #6
    You know that [tex] \tan x = \frac{\sin x}{\cos x} [/tex] and [tex] \cot x = \frac{\cos x}{\sin x} [/tex]


    [tex] \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^{2} x + \cos^{2} x}{\sin x \cos x} [/tex].

    [tex] \sin^{2} x + \cos^{2} x = 1 [/tex] so we have [tex] \frac{1}{\sin x \cos x} [/tex].

    Now we know the identity [tex] \sin 2x = 2\sin x \cos x [/tex]. Thus [tex] \sin x \cos x = \frac{1}{2}\sin 2x [/tex].

    So now we have [tex] \frac{1}{\frac{1}{2}\sin 2x} [/tex]. Remembering that [tex] \frac{1}{\sin x} = \csc x [/tex] we can deduce that [tex] \frac{1}{\frac{1}{2}\sin 2x} = 2\csc 2x [/tex]
    Last edited: Dec 2, 2006
  8. Dec 2, 2006 #7
    isnt it supposed to equal 2csc2x not 2cscx?
  9. Dec 2, 2006 #8


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    It does he probably made a typo. There are loads of ways to do it but basically all you have to do is find a page full of trig identities and fiddle about.
  10. Dec 5, 2006 #9
    Is it incorrect to work both sides of the problem and meet in the middle, or do you need to get one side to completely equal another, because I can easily get both sides to be 1/sinxcosx.
  11. Dec 5, 2006 #10
    You have to get one side completely equal to the other. Working both sides of the problem and "meeting in the middle" is not correct.
  12. Dec 5, 2006 #11

    Gib Z

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    And that is because in order to meet in the middle, your assuming the 2 sides are infact equal. If you mean change one side with identities, and change the other, and make them the same, then from there you can work backwards and get it from one side anyway.
  13. Nov 3, 2008 #12
    OK this is a easy problem

    L.S tanx+cotx

    = sinx^2+cos^2/cosxsinx


    R.S 2CSC2X
    =2*1/2sinxcosx (The 2's here cancel out)


    I tried to be as clear as possible, sorry but i'm not used to doing math on the computer :(
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