# Thread: Finding where a function = 0

1. ## Finding where a function = 0

I need to find where

20cos(5t) + 4sin(5t) = 0 for the first and second time after the initial t = 0

I get to

arctan (-5) = 5t

I can take arctan of -5 to get -1.3... then divide by 5, but that doesn't help me.

2. ## Re: Finding where a function = 0

Originally Posted by fourierT
I need to find where

20cos(5t) + 4sin(5t) = 0 for the first and second time after the initial t = 0

I get to

arctan (-5) = 5t

I can take arctan of -5 to get -1.3... then divide by 5, but that doesn't help me.

ok first rewrite this as a single trig function with a phase offset.

$20\cos(5t)+4\sin(5t)=A\cos(5t+\phi)$

$A=\sqrt{20^2+4^2}=\sqrt{416}$

$\phi = \arctan\left(\dfrac{4}{20}\right)\approx 0.197~ rad$

Now the cosine will be 0 for all odd multiples of $\dfrac{\pi}{2}$ so we look for

$5t+\phi=\dfrac{2k+1}{2}\pi$

$t=\dfrac{1}{5}\left(\dfrac{2k+1}{2}\pi-\phi\right)$

you can probably take it from here.

3. ## Re: Finding where a function = 0

Originally Posted by romsek
ok first rewrite this as a single trig function with a phase offset.

$20\cos(5t)+4\sin(5t)=A\cos(5t+\phi)$

$A=\sqrt{20^2+4^2}=\sqrt{416}$

$\phi = \arctan\left(\dfrac{4}{20}\right)\approx 0.197~ rad$

Now the cosine will be 0 for all odd multiples of $\dfrac{\pi}{2}$ so we look for

$5t+\phi=\dfrac{2k+1}{2}\pi$

$t=\dfrac{1}{5}\left(\dfrac{2k+1}{2}\pi-\phi\right)$

you can probably take it from here.
$t=\dfrac{1}{5}\left(\dfrac{3}{2}\pi-\phi\right)$

and

$t=\dfrac{1}{5}\left(\dfrac{5}{2}\pi-\phi\right)$

Are these correct?

4. ## Re: Finding where a function = 0

Your first way of attacking it would have worked but you goofed somewhere.

$20\cos(5t)+4\sin(5t)=0$

$20\cos(5t)=-4\sin(5t)$

$-5=\tan(5t)$

ah I see what happened. You just took the arctan here ignoring the fact that the tangent function is periodic. If you do this

$-5 = \tan(5t+k\pi)$ and solve for $k=0,1$ you'll arrive at the same answers.

5. ## Re: Finding where a function = 0

Originally Posted by romsek
Your first way of attacking it would have worked but you goofed somewhere.

$20\cos(5t)+4\sin(5t)=0$

$20\cos(5t)=-4\sin(5t)$

$-5=\tan(5t)$

ah I see what happened. You just took the arctan here ignoring the fact that the tangent function is periodic. If you do this

$-5 = \tan(5t+k\pi)$ and solve for $k=0,1$ you'll arrive at the same answers.
Wouldn't I solve for k(1,2) because it asks for where it = 0 after the initial t =0?

6. ## Re: Finding where a function = 0

yes, you're correct.

7. ## Re: Finding where a function = 0

Originally Posted by romsek
yes, you're correct.