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Homework Help: Need some help: Trig Identities

  1. May 5, 2008 #1
    Theres a few...
    Write each expression as a single trigonometric ratio or as the number 1.


    1) sint+(cott)(cost)

    2) (sec x)(sin^2x)(csc x)


    For number one I went like this:
    sin t + ((1/cot)(cos/1))
    sin t + (cos t/cot t)
    sin t + (cos t/1)( sin t/cos x)
    (sin t cos t)/1 + (sin t cot t)/1

    But then I get stuck.


    For number 2 I went like this:
    (secx)(sin^2x)(cscx)
    (1/cosx)(sin^2x/1)(1/sinx)
    sin^2x/(cosx)(sinx)

    But then I got stuck again. :confused:
     
  2. jcsd
  3. May 5, 2008 #2

    rock.freak667

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    sint+(cott)(cost)

    [tex]= sint + \frac{cos^2 t}{sint} = \frac{?+??}{sint}[/tex]


    [tex]secx*sin^2x*cosecx=\frac{1}{cosx}\times sin^2x \times \frac{1}{sinx}[/tex]

    [itex]sin^2x \times \frac{1}{sinx}[/itex] gives what? and then that times [itex]\frac{1}{cosx}[/itex] gives what?
     
  4. May 5, 2008 #3

    CompuChip

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    Err, what is the question? If I read your notation correctly, first you say it is
    and then you proceed to calculate
     
  5. May 5, 2008 #4
    Yes, the above statements are right. Your mistake lies in the fact that you mean to write 1/ tan t , and not 1/cott to represent cott.
     
  6. May 6, 2008 #5
    would it be:

    (sin^2x) x (1/sinx) = sin^2x/sinx

    (sin^2x/sinx) x (1/cosx) = (sin^2x)/(sinx cosx)?
     
  7. May 6, 2008 #6
    You are correct, but rock.freak is showing you that you can simplify sin2x/sinx. Hint: what is y2/y, or 52/5?
     
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