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Thread: trig identity

  1. #1
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    trig identity

    hi

    have this function g(x)=cos(x)exp(2/3sin(x)) 0<x<2pie

    also have derivative $\displaystyle g'(x)=1/3(2-3sinx-2sin^2x)$(exp2/3sinx)

    i need to find the stationary points of g(x) and classify each point min and max. the domain is [0,2pie]

    my main problem is i do not know how to get any values to use in original function i am not sure about trig identity. i am sure that i need to set $\displaystyle 1/3(2-3sinx-2sin^2x)$ to 0 but thats as far i as go.

    any help would be great
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by smartcar29 View Post
    hi

    have this function g(x)=cos(x)exp(2/3sin(x)) 0<x<2pie

    also have derivative $\displaystyle g'(x)=1/3(2-3sinx-2sin^2x)$(exp2/3sinx)

    i need to find the stationary points of g(x) and classify each point min and max. the domain is [0,2pie]

    my main problem is i do not know how to get any values to use in original function i am not sure about trig identity. i am sure that i need to set $\displaystyle 1/3(2-3sinx-2sin^2x)$ to 0 but thats as far i as go.

    any help would be great
    Don't worry about the $\displaystyle \frac{1}{3}$; it's just a constant multiple. To solve $\displaystyle 2-3\sin x-2\sin^2x=0$, just treat $\displaystyle \sin x$ like you would $\displaystyle x$ and factor it.

    We know that $\displaystyle 2-3x-2x^2$ factors into $\displaystyle (1-2x)(2+x)$, so $\displaystyle 2-3\sin x-2\sin^2x$ factors into $\displaystyle (1-2\sin x)(2+\sin x)$. Then set each part equal to 0.

    Spoiler:
    $\displaystyle 1-2\sin x=0 \implies \sin x =\frac{1}{2} \implies x = \frac{\pi}{6}$ or $\displaystyle \frac{5\pi}{6}$

    $\displaystyle 2+\sin x=0 \implies \sin x=-2$ which has no solutions.

    So at $\displaystyle x=\frac{\pi}{6}$ and $\displaystyle x=\frac{5\pi}{6}, g'(x)=0$.
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