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Math Help - Derivatives + Trig

  1. #1
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    Derivatives + Trig

    The hypotenuse of a right-angled triangle is 12cm in length. Find the measure of the angles in the triangle that maximize its perimeter.

    This is my function for perimeter that I came up with

    12 + 12sin(theta)+12cos(theta)

    Derivative

    12cos(theta)-12sin(theta)

    Are you just supposed to know when cos(theta) = sin(theta) or is there a way of figuring it out in different situations?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    The hypotenuse of a right-angled triangle is 12cm in length. Find the measure of the angles in the triangle that maximize its perimeter.

    This is my function for perimeter that I came up with

    12 + 12sin(theta)+12cos(theta)

    Derivative

    12cos(theta)-12sin(theta)

    Are you just supposed to know when cos(theta) = sin(theta) or is there a way of figuring it out in different situations?
    cos(x)=sin(x) at pi/4..think arctan(1)
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  3. #3
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    think arctan(1)
    Have not learned/Don't know what it is.

    So is there any way other than "think arctan"..?
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  4. #4
    o_O
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    Just to be a bit more explicit:
    \frac{\cos \theta}{{\color{blue}\cos \theta}} = \frac{\sin \theta}{{\color{blue}\cos \theta}}
    1 = \tan \theta
    \theta = \tan^{-1} (1)

    \arctan(x) is just another notation for \tan^{-1}(x)
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by theowne View Post
    The hypotenuse of a right-angled triangle is 12cm in length. Find the measure of the angles in the triangle that maximize its perimeter.

    This is my function for perimeter that I came up with

    12 + 12sin(theta)+12cos(theta)

    Derivative

    12cos(theta)-12sin(theta)

    Are you just supposed to know when cos(theta) = sin(theta) or is there a way of figuring it out in different situations?
    What we have is

    P(\theta)=12+12\sin(\theta)+12\cos(\theta)

    \frac{dP}{d\theta}=12\cos(\theta)-12\sin(\theta)

    setting equal to zero and solving gives

    0=12\cos(\theta)-12\sin(\theta) \iff 12\sin(\theta)=12\cos(\theta)=
     \sin(\theta)=\cos(\theta) \iff \tan(\theta) =1 \iff \theta=\tan^{-1}(1) =\frac{\pi}{4}
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  6. #6
    Moo
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    Hello,

    When is \cos(\theta)=\sin(\theta) ?

    By using this identity : \cos^2(x)+\sin^2(x)=1, \sin(x)=\sqrt{1-\cos^2(x)}

    Hence, we want to solve for \theta in \cos(\theta)=\sqrt{1-\cos^2(\theta)}

    Square the equation :

    \Longrightarrow \cos^2(\theta)=1-\cos^2(\theta)

    \Longrightarrow 2\cos^2(\theta)=1

    \Longrightarrow \cos^2(\theta)=\frac 12

    \Longrightarrow \cos(\theta)=\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2}

    -------> \theta=\dots



    I did this because you said you didn't want to "think arctan"... ^^'
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