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Homework Help: Another Trig. Identity Question Any takers?

  1. Nov 7, 2004 #1
    1.) 1/(csc x)(sec^2 x) = ?

    a.) 1/(sin x)(cos^2 x)
    b.) sin x - sin^3 x
    c.) 1/(sin x)(1+tan^2 x)
    d.) sin x - (1/1+tan^2 x)
    e.) 1+tan^3 x



    Here's my work: 1/(csc x)(tan^2 x + 1)
    (1/sin x) * ((cos x/sin x)(cos x/sin x) + 1)

    So, am I on the right track here? Or do I need to do completely different stuff? I can't seem to find any of those 5 answers no matter what I do...
     
  2. jcsd
  3. Nov 7, 2004 #2
    For (1), rewrite the equation in terms of sine and cosine:

    [tex]\frac{1}{\csc{x}\sec^2{x}}&=\sin{x}\cos^2{x}[/tex]

    And take it from there (hint: use the basic Pythagorean identity).
    However, I'm assuming you meant that original equation is
    [tex]\displaystyle{\frac{1}{\csc{x}\sec^2{x}}}[/tex].
     
    Last edited: Nov 7, 2004
  4. Nov 7, 2004 #3

    kreil

    User Avatar
    Gold Member

    1/(cscx)(sec^2x)
    1/(1/sinxcos^2x)

    mult. top and bottom by sinxcos^2x

    sinxcos^2x

    Remember that cos^2x=1-sin^2x
    sinx(1-sin^2x)
    sinx-sin^3x

    so the answer is (b)
     
  5. Nov 7, 2004 #4
     
  6. Nov 7, 2004 #5
    Right, that's what I was trying to lead math_fortress to. ;)
     
  7. Nov 8, 2004 #6
    ahh my bad.. im sorry...
     
  8. Nov 8, 2004 #7
    No problem. :)
     
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