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Math Help - something's not working

  1. #1
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    something's not working

    I have three problems, and none of them seem to work. they give us a hint to use trig identities.

    48) dy/dx = sec^4 (x)

    50) dy/dx = 4 cos^2 (x)

    52) dy/dx = (cos^4 (x) - sin^4 (x))

    I have no idea how to do any of these. I spent the last 50 minutes trying
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  2. #2
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    Quote Originally Posted by jahichuanna View Post
    I have three problems, and none of them seem to work. they give us a hint to use trig identities.

    48) dy/dx = sec^4 (x)

    50) dy/dx = 4 cos^2 (x)

    52) dy/dx = (cos^4 (x) - sin^4 (x))

    I have no idea how to do any of these. I spent the last 50 minutes trying
    48)  \int \sec^4 x \,dx = \int \sec^2x \sec^2x \, dx let u = \tan x
    50)  \cos^2x = \frac{1 + \cos 2x}{2}

    52) factor
    \cos^4 x - \sin^4 x = \left( \cos^2x - \sin^2x\right) \left( \cos^2x + \sin^2x\right) = \cos 2x
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  3. #3
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    \sec^4{x} = \sec^2{x} \cdot \sec^2{x} = (1 + \tan^2{x})\sec^2{x}

    see u and du ?


    4\cos^2{x} = 4\left(\frac{1+\cos(2x)}{2}\right) = 2 + 2\cos(2x)


    \cos^4{x} - \sin^4{x} = (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x} = \cos(2x)


    study your identities.
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  4. #4
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    \cos^4{x} - \sin^4{x} = (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x} = \cos(2x)


    study your identities.[/quote]


    I don't see how you got that math.....

    (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x}
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  5. #5
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    Quote Originally Posted by jahichuanna View Post
    \cos^4{x} - \sin^4{x} = (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x} = \cos(2x)


    study your identities.

    I don't see how you got that math.....

    (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x}[/quote]

    which part do you not "see"?
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  6. #6
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    Quote Originally Posted by skeeter View Post
    I don't see how you got that math.....

    (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x}

    wouldnt that work out to be

    (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^4{x} - \sin^4{x}?
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  7. #7
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    Gotta remember that Cos2(x)+sin(2)x = 1
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  8. #8
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    Quote Originally Posted by Slipery View Post
    Gotta remember that Cos2(x)+sin(2)x = 1
    oh yeah.......ur right. i do need to study these.
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