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Prove the trig identity

  1. Dec 1, 2009 #1
    csc(theta) - sin(theta) = cos(theta)*cot(theta)

    I'm supposed to write a proof for this but to be honest I'm not really sure where I should even start. The prof taught to take one side of the equation and simply manipulate each part into its equivalent until the other side of the equation was reached (if that makes sense). I just can't seem to get it flowing.
     
  2. jcsd
  3. Dec 1, 2009 #2
    Some work I'm trying,

    I know that csc(theta) = 1/sin(theta). But if I substitute for this I simply get [1/sin(theta)] - sin(theta). I'm just really stuck and could use a start as to how to go about proving this.
     
  4. Dec 1, 2009 #3
    actually this identity isn't that challenging

    as you said csc = 1/sin *i'm not gonna keep saying theta, you should know it's there

    so then you'll have 1/sin - sin, find common denominator by just multiplying sin by sin/sin to get (1 - sin^2)/sin

    1-sin^2 is just cos^2, so know you have (cos^2)/sin, which is just cos(cot)

    understand?
     
  5. Dec 1, 2009 #4
    Muliply both sides by some thing that will make it look more familiar.
     
  6. Dec 1, 2009 #5
    Okay guys, that helps me out alot. I guess I was on the right track I'm just too out of it to realize that I needed to multiply to get a common denominator. Thanks again for the guidance
     
  7. Dec 1, 2009 #6
    no problem all it takes is some manipulation and knowledge of other identities to get these
     
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