SOLUTION: I read through Question 869777, but did not understand the response given to it, is there perhaps another way of wording it? I appreciate the answer supplied, but can't seem to wra

Algebra ->  Trigonometry-basics -> SOLUTION: I read through Question 869777, but did not understand the response given to it, is there perhaps another way of wording it? I appreciate the answer supplied, but can't seem to wra      Log On


   



Question 870024: I read through Question 869777, but did not understand the response given to it, is there perhaps another way of wording it? I appreciate the answer supplied, but can't seem to wrap my head around the concept
Can you help me to solve the following question:
The vertical displacement of the end of a robot arm (in mm) at time t (in seconds) is given by
y=1+4cos(2t)-4sin(4t)
a) find all times, t>0 (in exact form i.e. in terms of pi) where the vertical displacement is 1mm, i.e. y=1
b) what is the first time, t>0 that the vertical displacement is 1mm? give you answer exactly and to 2 decimal places?

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
y=1%2B4cos%282t%29-4sin%284t%29=1%2B4%22%5B%22cos%282t%29-sin%284t%29%22%5D%22 is a periodic function, waving above and below y=1 , whose graph looks like this:
graph%28600%2C300%2C-1.6%2C6.4%2C-7%2C9%2C1%2B4cos%282x%29-4sin%284x%29%2C1%29 Its period is pi because at t and t%2Bpi the function has the same y value:
y=1%2B4cos%282t%2B2pi%29-4sin%284t%2B4pi%29=y=1%2B4cos%282t%29-4sin%284t%29 .
The function will take the value y=1 an infinite number of times.
For an answer to part a) we can list the first few in order and then write ", ..." or we can write it as a formula.
Either way, it is not a simple answer.

The double angle trigonometric identity
sin%282A%29=2sin%28A%29%2Acos%28A%29 can be applied, with A=2t to get
y=1%2B4%22%5B%22cos%282t%29-2sin%282t%29cos%282t%29%22%5D%22
Taking out cos%282t%29 as a common factor, we get
y=1%2B4%22%5B%22cos%282t%29%281-2sin%282t%29%29%22%5D%22
y=1%2B4%2Acos%282t%29%2A%281-2sin%282t%29%29
When will we have y=1 ?
We have to solve
1=1%2B4%2Acos%282t%29%2A%281-2sin%282t%29%29<--->4%2Acos%282t%29%2A%281-2sin%282t%29%29=0
That will be true when one of those factors (either cos%282t%29 or %281-2sin%282t%29%29 ) is zero.

cos%282t%29=0 --> 2t=pi%2F2 or 2t=3pi%2F2 or anything you can get by adding 2pi to a previous answer.
So t=pi%2F4 or t=3pi%2F4 or anything you can get by adding pi to a previous answer.

1-2sin%282t%29=0<--->sin%282t%29=1%2F2<--->2t=pi%2F6 or 2t=5pi%2F6 or anything you can get by adding 2pi to a previous answer.
That means t=pi%2F12 or t=5pi%2F12 or anything you can get by adding pi to a previous answer.

a) The list of times when y=0 is
pi%2F12%22%2C%22 pi%2F4=3pi%2F12%22%2C%22 5pi%2F12%22%2C%22 3pi%2F4=9pi%2F12%22%2C%22 13pi%2F12=pi%2F12%2Bpi%22%2C%22%22...%22 (and keep adding pi to previous answers).
You could state that the answers are
t=%28n%2B1%29pi%2F12 , t=%28n%2B3%29pi%2F12 , t=%28n%2B5%29pi%2F12 , and t=%28n%2B9%29pi%2F12 , for any non-negative integer n .
I could not think of a simple and elegant way to express that as one formula.

b) The first time y=1 happens when highlight%28t=pi%2F12=0.26%29