# Solve for t

• Dec 3rd 2012, 03:28 PM
sakonpure6
Solve for t
I have the following equation $y=-20cos225t+50$ 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:

$60=-20cos225t+50$
$t=10/-20cos225$
$t= root 2/2$
• Dec 3rd 2012, 03:39 PM
skeeter
Re: Solve for t
is the equation ...

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

or

$y = -20\cos(225) \cdot t + 50$
• Dec 3rd 2012, 03:42 PM
sakonpure6
Re: Solve for t
Quote:

Originally Posted by skeeter
is the equation ...

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

or

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

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

if it's the other equation then root2/2 works... so what did i do wrong?
• Dec 3rd 2012, 04:01 PM
skeeter
Re: Solve for t
Quote:

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

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

the whole thing ...

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

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

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

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

if $t$ is in radians ...

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

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

if $t$ is in degrees ...

$60 = 225t$

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

note that these are the principle solutions ... there are more.
• Dec 3rd 2012, 04:22 PM
Soroban
Re: Solve for t
Hello, sakonpure6!

Quote:

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

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

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

. . . . $=\;\text{-}20\left(\text{-}\tfrac{1}{2}\right) + 50 \;=\;10 + 50 \;=\;60$
• Dec 3rd 2012, 04:27 PM
sakonpure6
Re: Solve for t
Quote:

Originally Posted by Soroban
Hello, sakonpure6!

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

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

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

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, 04:56 PM
skeeter
Re: Solve for t
my mistake ...

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

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

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

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

if $t$ is in radians ...

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

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

if $t$ is in degrees ...

$120 = 225t$

$t = \frac{120}{225} = \frac{8}{15}$
• Dec 3rd 2012, 04:57 PM
sakonpure6
Re: Solve for t
alright thank you very much!