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Math Help - Basic trigonometric identities problem?

  1. #1
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    Basic trigonometric identities problem?

    Can anyone show me how to do this problem?
    1+cot^2x-cos^2x-cos^2xcotx^2
    This if how far I have gotten

    cscx^2-cos^2x-cos^2x (cosx^2/sinx^2)

    (1/sinx^2)-cos^2x-cos^2x(cosx^2/sinx^2)

    I am not sure how to proceed unfortunately?
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  2. #2
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    What are you trying to show that it is equivalent to?
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    I am supposed to simplify it as much as possible.
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    Get everything to have a common denominator of \displaystyle \sin^2{x} and simplify the numerator...
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    Quote Originally Posted by homeylova223 View Post
    Can anyone show me how to do this problem?
    1+cot^2x-cos^2x-cos^2xcotx^2
    1+\cot^2{x}-\cos^2{x}-\cos^2{x}\cot^2{x}

    1+\cot^2{x}-\cos^2{x}(1 + \cot^2{x})

    (1+\cot^2{x})(1 - \cos^2{x})

    \csc^2{x} \cdot \sin^2{x} = 1
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    I have another trigonometric identities I need some help in although I do not want to start a new thread. It is similar.

    cos^4x+2cos^2xsin^2x+sin^4x I need to simplify it this is what I have done thus far
    Factor cos^2x
    cos^2x+2sin^2x+sin^4x factor out sin^2x
    then I get this cos^2x+2+sin^2x But I do not think this is right?
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  7. #7
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    Quote Originally Posted by homeylova223 View Post
    I have another trigonometric identities I need some help in although I do not want to start a new thread. It is similar.

    cos^4x+2cos^2xsin^2x+sin^4x I need to simplify it this is what I have done thus far
    Factor cos^2x
    cos^2x+2sin^2x+sin^4x factor out sin^2x
    then I get this cos^2x+2+sin^2x But I do not think this is right?
    \cos^4{x}+2\cos^2{x}\sin^2{x}+\sin^4{x} = (\cos^2{x} + \sin^2{x})^2 = ???
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  8. #8
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    Oh it is one.
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