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    john1999's Avatar
    john1999 Posts: 14, Reputation: 1
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    #1

    May 24, 2012, 06:58 AM
    simplify the following
    y(x)=4/(1+tan^2x)-3
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    ebaines Posts: 12,131, Reputation: 1307
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    #2

    May 24, 2012, 07:03 AM
    Quote Originally Posted by john1999 View Post
    y(x)=4/(1+tan^2x)-3
    Hint: replace tan^2x with sin^2x/cos^2x and see where it takes you.
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    john1999 Posts: 14, Reputation: 1
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    #3

    May 24, 2012, 07:14 AM
    Quote Originally Posted by ebaines View Post
    Hint: replace tan^2x with sin^2x/cos^2x and see where it takes you.
    Didn't help
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    #4

    May 24, 2012, 07:24 AM
    Quote Originally Posted by john1999 View Post
    didnt help
    What did you get? After making this substitution try multiplying numerator and denominator by cos^2x.
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    john1999 Posts: 14, Reputation: 1
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    #5

    May 24, 2012, 07:29 AM
    I know to identities this has just thrown me to the deep end
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    #6

    May 24, 2012, 07:46 AM
    I got it down to y(x)=4cos^2/(1+(1/cos^2x)-3
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    #7

    May 24, 2012, 08:01 AM
    no one can help?. I put it in wolfram and it gives y(x)=1-4sin^2(x)

    any help on how to get to there
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    #8

    May 24, 2012, 08:20 AM
    Quote Originally Posted by john1999 View Post
    i got it down to y(x)=4cos^2/(1+(1/cos^2x)-3
    I think you made a mistake - what happened to the sin^2x term in the denominator?

    Substituting :



    Now multiply both numerator and denomninator by cos^2x:



    You should recognize that , so this becomes . Now if you want you can subsitite to get the Wolfram solution.
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    john1999 Posts: 14, Reputation: 1
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    #9

    May 24, 2012, 08:34 AM
    yeah my mistake was when I multiply sin^2x/cos^2x by cos^2x would not that cancel out the cos on the denominator?

    what program you use to write your equations?
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    #10

    May 24, 2012, 08:37 AM
    Quote Originally Posted by john1999 View Post
    yeah my mistake was when i multiply sin^2x/cos^2x by cos^2x would not that cancel out the cos on the denominator?
    yes it does, leaving the sin^2x term..

    Quote Originally Posted by john1999 View Post
    what program you use to write your equations?
    LaTeX - it allows for embedding math equations into questions and responses. You can learn about it here:
    https://www.askmehelpdesk.com/math-s...las-50415.html
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    john1999 Posts: 14, Reputation: 1
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    #11

    May 24, 2012, 08:44 AM
    yeah agreed so for my information in the denominator

    the sin^2x/cos^2x was multiply by cos^2x/cos^2x to leave sin^2x which makes it

    4/1+sin^2x

    then multiply that again by cos^2x/cos^2x

    to get 4cos^2x/1+sin^2x+cos^2x

    in your example where did the +1 go
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    #12

    May 24, 2012, 08:48 AM
    Where did the 1+ go on the bottom?
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    #13

    May 24, 2012, 09:05 AM
    ?
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    #14

    May 24, 2012, 11:20 AM
    Quote Originally Posted by john1999 View Post
    yeah agreed so for my information in the denominator

    the sin^2x/cos^2x was multiply by cos^2x/cos^2x to leave sin^2x which makes it

    4/1+sin^2x
    No - what I did was this (leaving the -3 term off for simplicity):



    I hope this is clear.

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