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Homework Help: Trigonometry Refresher

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm currently in cal 1 and we are doing trigonometry identities for review. I can't recall ever seeing a sum/difference formula for secant in precal

    (1) Prove secant² (4θ) / 2 - secant² (4θ) = secant (8θ)
    (2) Prove tan(x)/(1-cot(x)) + cot(x)/(1-tan(x)) = sec(x)csc(x) + 1


    3. The attempt at a solution

    (1) secant² (4θ) / 2 - secant² (4θ) = -1/2 *(sec²(4θ))

    (2) tan(x)/(1-cot(x)) + cot(x)/(1-tan(x))

    =[ tan(x)(1-tan(x)) + cot(x)(1-cot(x))] /(1-cot(x))(1-tan(x))

    = [tan(x) - tan²(x) + cot(x) - cot²(x)]/ (1-tan(x)-cot(x)+cot(x)tan(x))

    = sin(x)/cos(x) - sin²(x)/cos²(x) + cos(x)/sin(x) - cos²(x)/sin²(x) / 2-tan(x)-cot(x)

    = [sin3(x)cos(x)-sin4(x) +cos3(x)sin(x)-cos4(x)]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

    = [sin3(x)cos(x) + cos3(x)sin(x) - (sin4(x) + cos4(x))]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

    = [sin(x)cos(x)[sin2(x)+cos2(x)] -1]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

    = sin(x)cos(x) -1 / cos²(x)sin²(x)(2-sin(x)/cos(x)-cos(x)/sin(x))

    = sin(x)cos(x) -1/ 2cos²(x)sin²(x) - sin3(x)cos(x) -cos3(x)sin(x)

    = sin(x)cos(x) -1/ 2cos²(x)sin²(x) - sin(x)cos(x)(sin²(x) + cos²(x))

    = sin(x)cos(x) -1 /sin(x)cos(x)(2sin(x)cos(x)-1)

    = 1/2*csc(x)sec(x)


    This is where I'm stuck I can't find a way for it to look like what is trying to be proven here
     
    Last edited: Nov 8, 2008
  2. jcsd
  3. Nov 8, 2008 #2
    (1) I think you have to combine secant² (4θ) / 2 - secant² (4θ) into one fraction by making it (secant² (4θ) - 2secant² (4θ) ) / 2 then using double angle or half angle identity.

    Your first line for (2) doesn't comply with me. It's like you multiplied by (1-tan(x)) for the first term and your (1-cot(x)) on its denominator disappeared.
     
  4. Nov 8, 2008 #3
    Hmmm...I think I'll just figure this out on my own. There should have been more brackets in the post to lessen the confusion but I doubt anyone is going to volunteer to solve it
     
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