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December 23rd, 2017, 05:28 AM   #1
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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.

Last edited by skipjack; December 23rd, 2017 at 09:27 AM.
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December 23rd, 2017, 06:13 AM   #2
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$\sin\theta$ is periodic.
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December 23rd, 2017, 09:30 AM   #3
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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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