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 December 22nd, 2017, 02:10 AM #1 Newbie   Joined: May 2017 From: uk Posts: 9 Thanks: 0 Coordinates where gradient is zero of trig functions Hi everyone, I have two questions I am struggling with. They are: work out the coordinates when the gradient of these functions is equal to zero t=cos(theta) and Z=sin(2theta) I think I need to find the derivatives, but then what? Thanks in advance; have a nice day. Last edited by skipjack; December 23rd, 2017 at 08:07 AM.
December 22nd, 2017, 06:39 AM   #2
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Quote:
 Originally Posted by DomB Hi everyone, I have two questions I am struggling with. They are: work out the coordinates when the gradient of these functions is equal to zero t=cos(theta) and Z=sin(2theta) I think I need to find the derivatives, but then what?
Set each derivative equal to zero and solve for $\theta$.

Last edited by skipjack; December 23rd, 2017 at 08:08 AM.

 December 23rd, 2017, 04:07 AM #3 Newbie   Joined: May 2017 From: uk Posts: 9 Thanks: 0 Think I just get 0,0 by doing that. Last edited by skipjack; December 23rd, 2017 at 08:08 AM.
December 23rd, 2017, 07:46 AM   #4
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 Originally Posted by DomB Think I just get 0,0 by doing that.
$t =\cos{\theta}$

$\dfrac{dt}{d\theta} = -\sin{\theta} = 0 \implies \theta = k\pi \, , \, k \in \mathbb{Z}$

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$z = \sin(2\theta)$

$\dfrac{dz}{d\theta} = 2\cos(2\theta) = 2(2\cos^2{\theta} - 1) = 0 \implies \cos{\theta} = \pm \dfrac{1}{\sqrt2}$

so, $\theta = \, ?$

Last edited by skipjack; December 23rd, 2017 at 08:11 AM.

 December 23rd, 2017, 08:19 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,526 Thanks: 1750 $\cos(2\theta) = 0$ directly implies that $2\theta = \pi/2 + k\pi$.
February 15th, 2018, 11:13 AM   #6
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Quote:
 Originally Posted by DomB Think I just get 0,0 by doing that.
Are you just finding arcsin(0) on your calculator? You need to know more about sin(x) than that to do these problems!

sin(x)= 0 for x any multiple of $\displaystyle \pi$.

cos(x)= 0 for x any odd multiple of $\displaystyle \frac{\pi}{2}$.

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