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 DomB December 23rd, 2017 04:28 AM

stationary points of a trig function

Hi, am I doing this right?

Find the coordinates when the gradient of this function is equal to zero

t = cos(θ)

f '(t) = -sin(θ)
$\ \ \ \ \ \ \ \$ = -sin(θ) = 0
$\ \ \ \$ θ = 0/-sin
$\ \ \ \$ θ = 0

Plug 0 in to original equation
t = cos(0)
t = 1
Coordinates = 0,1 where gradient is zero.

 greg1313 December 23rd, 2017 05:13 AM

$\sin\theta$ is periodic.

 skipjack December 23rd, 2017 08:30 AM

Having obtained -sin(θ) = 0, one cannot divide by "-sin".

Instead, solve sin(θ) = 0 by using knowledge of the sin function.

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