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Homework Help: Trigonometric Identities

  1. Dec 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Express (1+cot^2 x) / (cot^2 x) in terms of sinx and/or cosx

    2. Relevant equations
    cot(x) = 1/tan(x)
    sin^2(x) + cos^2(x) = 1

    3. The attempt at a solution
    I do not know if I am solving this problem correctly. Is there an easier route than the way I have solved it, if it is solved correctly?

    = (1+cot^2x) / (cot^2x)
    = 1+ [ (cos^2x) / (sin^2x) ] ÷ [ (cos^2x) / (sin^2x) ]
    = 1 + [ (cos^2x) / (sin^2x) ] x [ (sin^2x / cos^2x) ]
    = [ (sin^2x / sin^2x) + (cos^2x / sin^2x) ] x [ (sin^2x) / (cos^2x) ]
    = [ (sin^2x + cos^2x) / (sin^2x) ] x [ (sin^2x )/ (cos^2x) ]
    = [ 1 / sin^2x ] x [ sin^2x / cos^2x]
    = sin^2x / (sin^2x)(cos^2x)
    = 1 / cos^2x
     
  2. jcsd
  3. Dec 10, 2017 #2

    Dick

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    Looks just fine to me.
     
  4. Dec 10, 2017 #3

    Ray Vickson

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    Why don't you use the fact that
    $$
    \frac{1 +\cot^2 x}{\cot^2 x} = \frac{1}{\cot^2 x} + 1 ?
    $$
    Then you can finish off the whole thing in one more line of simple algebra (plus the identity ##\cos^2 x + \sin^2 x = 1##).
     
  5. Dec 10, 2017 #4
    so if 1 / cot^2x = tan^2x = sin^2x / cos^2x

    Then all we have to do is: = (sin^2x/cos^2x ) + (cos^2x/cos^2x)
    =( sin^2x + cos^2x)/ cos^2x
    = 1 / cos^2x

    Thanks so much haha, I didn't even notice that. Also another question for proofs. We know that sin^2x + cos^2x = 1 (pythagorean identity). Can we say the same for sinx + cosx= 1? I am pretty sure not but just double checking.
     
  6. Dec 10, 2017 #5

    Dick

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    Absolutely not. Try it if ##x=\pi##.
     
  7. Dec 31, 2017 #6
    The brackets [ ] on your second line are incorrect
     
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