We need to get this in a simpler form. In this one, notice that the angles in the brackets are not the same!

We must simplify them first so the angles in the brackets are the same.

`i=2\ sin(t-pi/3)-cos(t+pi/2)`

`=2(sin\ t\ cos\ pi/3-cos\ t\ sin\ pi/3)` `-(cos\ t\ cos\ pi/2-sin\ t\ sin\ pi/2)`

`=2((sin\ t)/2-0.866\ cos\ t)` `-(cos\ txx0-sin\ t xx1)`

`=sin\ t-1.732\ cos\ t+sin\ t`

`=2\ sin\ t-1.732\ cos\ t`

Now we can express i in the form R sin(t + α).

`R=sqrt(2^2+1.732^2)=2.646`

`alpha=arctan(1.732/2)=0.714\ text(radians`

So

`2\ sin\ t − 1.732\ cos\ t = 2.646\ sin(t − 0.714)`

So the maximum value of this is `2.646\ "A"`.

To find when this occurs, we need to solve:

`2.646\ sin(t − 0.714) = 2.646`

i.e. `sin(t − 0.714) = 1`

`t − 0.714 = π/2`

`t = 2.29`

So `t = 2.29\ "s"` is the time when the maximum is first reached.