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.