Note that normally we take t ≥ 0. For this sort of example, we must use radians for the angle.

We have:

`R=sqrt(12^2+5^2)=13` and `alpha=arctan(5/12)=0.39479`.

So `12\ sin t+5\ cos t=` `13\ sin(t+0.39479)`

So we can see that the amplitude is 13 A and this is the maximum value.

To find the first time that it occurs, we need to solve

`13\ sin(t+0.39479)=13`

That is

`sin(t+0.39479)=1`

Now sin θ = 1 for the first time when `theta=pi/2`. So we need to solve:

`t+0.39479=pi/2`

`t=pi/2-0.39479=1.176`

So the maximum value of 13 A will first occur at time t = 1.176 s.

We can see that this is correct from the graph:

`i=12\ sin t+5\ cos t`

Graph 12 sin t + 5 cos t