# Thread: Help with a trig question - Solving for variable

1. ## Help with a trig question - Solving for variable

Hello everyone, just having some trouble with my assignment.

I had to find the equation for velocity using the equation for vertical displacement so I did. The question asks to find the vertical displacement when the velocity is 0.8 m/s so as you can see I subbed it into the equation above. Problem is, im having trouble solving for 't'.

Can anyone help?... It would be much appreciated.

Ill need a lot of help for calculus so I hope I don't bother these forums!

2. $0.8= -1.2sin(2t)+0.8cos(t)$

Use double angle formula

$0.8= -1.2sin(t)cos(t)+0.8cos(t)$

Factor out cos(t)

$0.8= cos(t)(0.8-1.2sin(t))$

can you solve it from here?

3. Originally Posted by pickslides
$0.8= -1.2sin(2t)+0.8cos(t)$

Use double angle formula

$0.8= -1.2{\color{red}(2)}sin(t)cos(t)+0.8cos(t)$
$0.8= {\color{red}-2.4}sin(t)cos(t)+0.8cos(t)$

Factor out cos(t)

$0.8= cos(t)(0.8-{\color{red}2.4}sin(t))$

can you solve it from here?
Quoted with minor corrections.

01

4. Originally Posted by DrCheeky

Hello everyone, just having some trouble with my assignment.

I had to find the equation for velocity using the equation for vertical displacement so I did. The question asks to find the vertical displacement when the velocity is 0.8 m/s so as you can see I subbed it into the equation above. Problem is, im having trouble solving for 't'.

Can anyone help?... It would be much appreciated.

Ill need a lot of help for calculus so I hope I don't bother these forums!

Originally Posted by pickslides
$0.8= -1.2sin(2t)+0.8cos(t)$

Use double angle formula

$0.8= -1.2sin(t)cos(t)+0.8cos(t)$

Factor out cos(t)

$0.8= cos(t)(0.8-1.2sin(t))$

can you solve it from here?
I don't see how this (or its corrected incarnation) helps get an exact solution. It's 0.8 on the left hand side, not 0 ....

5. $0.8 = -1.2sin2t + 0.8cost$

$1 = -1.5sin2t + cost$(Divided everything by 0.8)

$1 = -1.5(2)sintcost + cost$ (Provided by you guys, correct?)

$1 = -3sintcost + cost$

$1 = cost (-3sint + 1)$(Factored out cost)

$0 = -3sint + 1$(cos^-1 on the other side)

$-1 = -3sint$

$t = sin^-1(-1/-3)$

$t = 0.34$

Can anyone see if thats correct?

6. Originally Posted by DrCheeky
$0.8 = -1.2sin2t + 0.8cost$

$1 = -1.5sin2t + cost$(Divided everything by 0.8)

$1 = -1.5(2)sintcost + cost$ (Provided by you guys, correct?)

$1 = -3sintcost + cost$

$1 = cost (-3sint + 1)$(Factored out cost)

$0 = -3sint + 1$(cos^-1 on the other side) No, you cannot do that!

$-1 = -3sint$

$t = sin^-1(-1/-3)$

$t = 0.34$

Can anyone see if thats correct?
No, not correct. See the red in the quote.

$\cos^{-1}(\cos t) = t$ (assuming that we are mindful of the domain and range), but $\cos^{-1}[\cos t (-3\sin t + 1)] \neq -3\sin t + 1$.

01

7. So what else can I do?

I'm 100% I got up to this part right so far (finding the derivative of vertical displacement equation and subbing in velocity). Cant find this t value!

8. Originally Posted by DrCheeky
So what else can I do?

I'm 100% I got up to this part right so far (finding the derivative of vertical displacement equation and subbing in velocity). Cant find this t value!

9. DrCheeky, you mentioned finding the derivative of the vertical displacement equation. What is it, out of curiosity?

01

10. Originally Posted by mr fantastic