# trig identity

• Sep 15th 2009, 09:41 AM
smartcar29
trig identity
hi

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

also have derivative $g'(x)=1/3(2-3sinx-2sin^2x)$(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 $1/3(2-3sinx-2sin^2x)$ to 0 but thats as far i as go.

any help would be great
• Sep 15th 2009, 10:07 AM
redsoxfan325
Quote:

Originally Posted by smartcar29
hi

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

also have derivative $g'(x)=1/3(2-3sinx-2sin^2x)$(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 $1/3(2-3sinx-2sin^2x)$ to 0 but thats as far i as go.

any help would be great

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

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

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

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

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