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Math Help - Maximum value

  1. #1
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    Question Maximum value

    Show that sinx(1+cosx) has a maximum value when x = \frac{\pi}{3}
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  2. #2
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    Hey!

    You need so show that

    f(x) := sin(x) [1 + cos(x)]

    f'(x) = 2(cos(x))^2 + cos(x) - 1

    f'(x) = 0 <=> x = -\frac{\pi}{3}, x = \frac{+\pi}{3} , x = ...

    f''(x) = - 4 sin(x) cos(x) - sin(x)

    f''(\frac{\pi}{3}) = -\frac{3*\sqrt{3}}{2} < 0

    => Maximum

    -----

    Is there any problem for you?

    Regards, Rapha
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  3. #3
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    Let f(x) = \sin{x}(1 + \cos{x})

    Differentiating,

    f'(x) = -\sin^2{x} + \cos{x}(1 + \cos{x})

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

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

    = 2\cos^2{x}  + \cos{x} - 1

    = (2\cos{x}  - 1)(\cos{x} + 1)

    Equate to zero and solve. For the x values you obtain, substitute back into f(x) and see which x value gives a higher y-value. You can now prove what is being asked.
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  4. #4
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    Trigonometry

    Hello varunnayudu
    Quote Originally Posted by varunnayudu View Post
    Show that sinx(1+cosx) has a maximum value when x = \frac{\pi}{3}
    You might find it easier to use the formula \sin 2x = 2\sin x \cos x before differentiating. This means you don't get any powers of 2. Instead you get:

    y = \sin x(1+\cos x)

    =\sin x + \frac{1}{2}\sin 2x

    \implies \frac{dy}{dx}=\cos x + \cos 2x

    =0 when

    \cos x = -\cos 2x

    Then use the fact that

    \cos A = -\cos B \implies A=\pi - B or \pi + B or 3\pi \pm B, etc.

    Taking the first one of these possibilities gives

    x = \pi - 2x

    \implies x=\frac{\pi}{3}

    Now differentiate \cos x + \cos 2x and show that this value of x gives a maximum value of y.

    Grandad
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