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Math Help - Derivative of this trig function

  1. #1
    Member Chokfull's Avatar
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    Arrow Derivative of this trig function

    Prove that

    \frac{d^n}{dx^n}(\sin^4x+\cos^4x)=4^{n-1}\cos(4x+\frac{n\pi}{2})

    The first thing to do, obviously, is to find the first derivative, which is

    4\sin^3x\cos x-4\cos^3x\sin x

    First I tried factoring, then using a couple of trigonometric identities.

     <br />
-4\sin x\cos x(\cos^2x-\sin^2x)=-4\sin x\cos x(\cos(2x))=-4\sin x\cos x(2\cos^2x-1)<br />

    None of this, however, gets me any closer to 4^{n-1}\cos(4x+\frac{n\pi}{2}). I differentiated a second time and got

    -4(\cos^4x+\sin^4x-24\sin^2x\cos^2x)

    I haven't double-checked my math on this last step but I'm pretty sure I'm right, but, regardless, it's not any closer either! So, I ask you, am i doing something wrong or just trying to follow the wrong steps?
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  2. #2
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    \displaystyle \sin^4{x}+\cos^4{x} = \frac{1}{4}\left(\cos{4x}+3\right), so f'(x) = -\sin{4x} = \cos\left(4x+\frac{\pi}{2}\right).
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  3. #3
    MHF Contributor Amer's Avatar
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    u can prove that by induction let

    \displaystyle P_k : \frac{d^k}{dx^k} (\sin ^4 x + \cos ^4 x ) = 4^{n-1} \cos \right(4x + \frac{k\pi}{2} \left)

    for k= 1

    -4\sin x \cos x ( \cos 2x ) = -2\sin 2x (\cos 2x ) = - \sin 4x
    left hand side

    4^0 \cos (4x + \frac{\pi}{2} ) = \cos 4x \cos \frac{\pi}{2} - \sin 4x \sin \frac{pi}{2} = - \sin 4x
    same as the left hand side

    now suppose P is true for n
    want P is true for k = n+1 try it
    \displaystyle P_n : \frac{d^n}{dx^n} (\sin ^4 x + \cos ^4 x ) = 4^{n-1} \cos \right(4x + \frac{n\pi}{2} \left)

    can u continue ?
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  4. #4
    Member Chokfull's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    \displaystyle \sin^4{x}+\cos^4{x} = \frac{1}{4}\left(\cos{4x}+3\right), so f'(x) = -\sin{4x} = \cos\left(4x+\frac{\pi}{2}\right).
    I don't get that first step...how do you get 1/4 (cos 4x+3) from sin^4 x+cos^4 x?
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by Chokfull View Post
    I don't get that first step...how do you get 1/4 (cos 4x+3) from sin^4 x+cos^4 x?
    Trig identities see here

    List of trigonometric identities - Wikipedia, the free encyclopedia
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  6. #6
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    Indeed...
    Quote Originally Posted by Chokfull View Post
    I don't get that first step...how do you get 1/4 (cos 4x+3) from sin^4 x+cos^4 x?
    \sin^4{x}+\cos^4{x} = \left(\sin^2{x}+\cos^2{x}\right)^2-2\left(\sin{x}\cos{x}\right)^2 = 1-2\left(\frac{1}{2}\sin{2x}\right)^2

    \displaystyle 1-\frac{1}{2}\sin^2{2x} =  \frac{2-\sin^2{2x}}{2} =  \frac{2-\left(\frac{1-\cos{4x}}{2}\right)}{2} = \frac{\cos{4x}+4-1}{4} = \frac{1}{4}\left(\cos{4x}+3\right).

    Or if you are comfortable/allowed to use tables at leisure, just add the two reductions from the list:

    \displaystyle \sin^4{x}+\cos^4{x} = \left(\frac{3 - 4 \cos{2x} + \cos{4x}}{8}\right)+\left(\frac{3 + 4 \cos{2x} + \cos{4x}}{8}\right) =  \frac{1}{4}\left(\cos{4x}+3\right).
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  7. #7
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    Derivative of this trig function

    Quote Originally Posted by Chokfull View Post
    Prove that

    \frac{d^n}{dx^n}(\sin^4x+\cos^4x)=4^{n-1}\cos(4x+\frac{n\pi}{2})

    The first thing to do, obviously, is to find the first derivative, which is

    4\sin^3x\cos x-4\cos^3x\sin x

    First I tried factoring, then using a couple of trigonometric identities.

     <br />
-4\sin x\cos x(\cos^2x-\sin^2x)=-4\sin x\cos x(\cos(2x))=\ \dots<br />

    Continuing from where I cut off your work ...

    = -2(\sin(2 x))(\cos(2x))

    = -\sin(4 x)

    Shifting \cos x by {\pi\over2} to the left gives -\sin x, so = -\sin(4 x)=\cos(4x+{\pi\over2}).

    So it works for n=1.

    Now, what if you take the derivative of \cos(4x+{\pi\over2}) with respect to x?
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  8. #8
    Member Chokfull's Avatar
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    thanks everyone
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