# Solve for t

• Dec 3rd 2012, 02:28 PM
sakonpure6
Solve for t
I have the following equation $\displaystyle y=-20cos225t+50$ If i set y= 60 and i solve i get root 2 over 2, howevef if i sub that back in the equation, i don't get 60 but I get about 68! Any help is appreciated. Here is what i did:

$\displaystyle 60=-20cos225t+50$
$\displaystyle t=10/-20cos225$
$\displaystyle t= root 2/2$
• Dec 3rd 2012, 02:39 PM
skeeter
Re: Solve for t
is the equation ...

$\displaystyle y = -20\cos(225t) + 50$

or

$\displaystyle y = -20\cos(225) \cdot t + 50$
• Dec 3rd 2012, 02:42 PM
sakonpure6
Re: Solve for t
Quote:

Originally Posted by skeeter
is the equation ...

$\displaystyle y = -20\cos(225t) + 50$

or

$\displaystyle y = -20\cos(225) \cdot t + 50$

This one $\displaystyle y = -20\cos(225t) + 50$

if it's the other equation then root2/2 works... so what did i do wrong?
• Dec 3rd 2012, 03:01 PM
skeeter
Re: Solve for t
Quote:

Originally Posted by sakonpure6
This one $\displaystyle y = -20\cos(225t) + 50$

if it's the other equation then root2/2 works... so what did i do wrong?

the whole thing ...

$\displaystyle 60 = -20\cos(225t) + 50$

$\displaystyle -10 = -20\cos(225t)$

$\displaystyle \frac{1}{2} = \cos(225t)$

$\displaystyle \cos^{-1}\left(\frac{1}{2}\right) = 225t$

if $\displaystyle t$ is in radians ...

$\displaystyle \frac{\pi}{3} = 225t$

$\displaystyle t = \frac{\pi}{675}$

if $\displaystyle t$ is in degrees ...

$\displaystyle 60 = 225t$

$\displaystyle t = \frac{60}{225} = \frac{4}{15}$

note that these are the principle solutions ... there are more.
• Dec 3rd 2012, 03:22 PM
Soroban
Re: Solve for t
Hello, sakonpure6!

Quote:

$\displaystyle y\:=\:-20(\cos225)t+50$
If I set $\displaystyle y= 60$ and I solve, I get $\displaystyle \tfrac{\sqrt{2}}{2}$
However if I sub that back in the equation, I don't get 60 but I get about 68! . How?

If $\displaystyle t = \tfrac{\sqrt{2}}{2}$, we have:

. . $\displaystyle \text{-}20(\cos225)\left(\tfrac{\sqrt{2}}{2}\right) + 50 \;=\;\text{-}20\left(\text{-}\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{\sqrt{2}}{ 2}\right) + 50$

. . . . $\displaystyle =\;\text{-}20\left(\text{-}\tfrac{1}{2}\right) + 50 \;=\;10 + 50 \;=\;60$
• Dec 3rd 2012, 03:27 PM
sakonpure6
Re: Solve for t
Quote:

Originally Posted by Soroban
Hello, sakonpure6!

If $\displaystyle t = \tfrac{\sqrt{2}}{2}$, we have:

. . $\displaystyle \text{-}20(\cos225)\left(\tfrac{\sqrt{2}}{2}\right) + 50 \;=\;\text{-}20\left(\text{-}\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{\sqrt{2}}{ 2}\right) + 50$

. . . . $\displaystyle =\;\text{-}20\left(\text{-}\tfrac{1}{2}\right) + 50 \;=\;10 + 50 \;=\;60$

I multiplied root2/2 by 225 before taking it's cosine, that was my mistake there and for skeeter how did you get -10? shouldnt it be 10
• Dec 3rd 2012, 03:56 PM
skeeter
Re: Solve for t
my mistake ...

$\displaystyle 60 = -20\cos(225t) + 50$

$\displaystyle 10 = -20\cos(225t)$

$\displaystyle -\frac{1}{2} = \cos(225t)$

$\displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = 225t$

if $\displaystyle t$ is in radians ...

$\displaystyle \frac{2\pi}{3} = 225t$

$\displaystyle t = \frac{2\pi}{675}$

if $\displaystyle t$ is in degrees ...

$\displaystyle 120 = 225t$

$\displaystyle t = \frac{120}{225} = \frac{8}{15}$
• Dec 3rd 2012, 03:57 PM
sakonpure6
Re: Solve for t
alright thank you very much!