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Trig geometry problem

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A triangle with x as the angle, 'a' as the adjacent, 8 as the opposite side and 'h' as the hypotenuse. An intermediate question was to show that sin(x)=cot(x). Now i have to write 'sin(x)=cot(x)' in terms of cos(x) only and solve the quadratic, stating correct value for cos(x)


    2. Relevant equations
    2. "sin(x)=cot(x)" in terms of cos(x). a*h=8^2


    3. The attempt at a solution
    3. I have used some trig identities and have got cos^2(x)+cos(x)-1=0. Is this correct? To solve i solved x^2+x-1 using quadratic formula and got 2 answers, one negative. So cos(x) is the positive value..?
     
  2. jcsd
  3. Sep 26, 2009 #2

    LCKurtz

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    How can you hope to "show" that sin(x) = cot(x) when that normally isn't true?

    If x was an angle such that sin(x) = cot(x), then cos(x) would satisfy your equation.
    Why couldn't cos(x) be negative? And what does all this have to do with your given triangle? What are you actually trying to do?
     
  4. Sep 27, 2009 #3
    For this triangle a*h=8^2. This can be shown by using a^2+b^2=c^2. To show that sin(x)=cot(x) 'in this case' : By using sin(x)=opp/hyp = 8/h and a*h=*^2. Rearrange for h=8^2/a

    so sin(x)=8/(8^2/a)=a/8 and cot(x)=adj/opp=a/8

    so sin(x)=cot(x) for this case.

    Yeah the angle is acute(sorry forgot to mention), so that means I am correct?
     
    Last edited: Sep 27, 2009
  5. Sep 27, 2009 #4
    Given what you wrote in the first post, h is the hypotenuse and a and 8 are the legs, so a2 + 82 = h2, not a*h = 82. Your substitution was wrong when rewriting sinx.

    sinx = a/h, cotx = a/8
     
  6. Sep 27, 2009 #5
    It might be helpful if i explain why a*h=8^2 for this triangle?

    h^2=8^2+a^2
    8=sqrt(h^2-a^2)
    8=sqrt(h^2-(h^2-8^2))
    8=sqrt(8^2)
    so 8=8
    so a*h=8^2



    I'm still needing conformation that I solved the correct quadratic
     
    Last edited: Sep 27, 2009
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