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Thread: Maximum value

  1. #1
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    Question Maximum value

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

    You need so show that

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

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

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

    $\displaystyle f''(x) = - 4 sin(x) cos(x) - sin(x)$

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

    => Maximum

    -----

    Is there any problem for you?

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

    Differentiating,

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

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

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

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

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

    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 $\displaystyle sinx(1+cosx)$ has a maximum value when $\displaystyle x = \frac{\pi}{3}$
    You might find it easier to use the formula $\displaystyle \sin 2x = 2\sin x \cos x$ before differentiating. This means you don't get any powers of 2. Instead you get:

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

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

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

    $\displaystyle =0$ when

    $\displaystyle \cos x = -\cos 2x$

    Then use the fact that

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

    Taking the first one of these possibilities gives

    $\displaystyle x = \pi - 2x$

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

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

    Grandad
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