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December 22nd, 2017, 02:10 AM   #1
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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.
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December 22nd, 2017, 06:39 AM   #2
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Quote:
Originally Posted by DomB View Post
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 $\theta$.
Thanks from DomB

Last edited by skipjack; December 23rd, 2017 at 08:08 AM.
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December 23rd, 2017, 04:07 AM   #3
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Think I just get 0,0 by doing that.

Last edited by skipjack; December 23rd, 2017 at 08:08 AM.
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December 23rd, 2017, 07:46 AM   #4
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Quote:
Originally Posted by DomB View Post
Think I just get 0,0 by doing that.
$t =\cos{\theta}$

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

------------------------------------------------------------------

$z = \sin(2\theta)$

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

so, $\theta = \, ?$

Last edited by skipjack; December 23rd, 2017 at 08:11 AM.
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December 23rd, 2017, 08:19 AM   #5
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$\cos(2\theta) = 0$ directly implies that $2\theta = \pi/2 + k\pi$.
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February 15th, 2018, 11:13 AM   #6
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Quote:
Originally Posted by DomB View Post
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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