My Math Forum stationary points of a trig function

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 December 23rd, 2017, 04:28 AM #1 Newbie   Joined: May 2017 From: uk Posts: 9 Thanks: 0 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 08:27 AM.
 December 23rd, 2017, 05:13 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond $\sin\theta$ is periodic.
 December 23rd, 2017, 08:30 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,054 Thanks: 1618 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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