Finding Range of Temperature (T) as t Approaches Infinity

[glid02]

I already found this equation:
T= 15/1.36(.6*sin(t)-cos(t))+100-23.971/e^(.6t)

Now I'm supposed to find the range of temperatures (T) as t approaches infinity.

I tried 100+(15/1.36*.6*sin(90))
and 100-(15/1.36*cos(0))

but that's not right. What am I missing?

Thanks.

 

[antonantal]

If I got it right your equation looks like this:
\frac{15}{1.36}[0.6sin(t)-cos(t)]+100-\frac{23.971}{e^{0.6t}}

Since \frac{23.971}{e^{0.6t}} goes to 0 when t goes to \infty you should calculate the maximum and the minimum of the expression \frac{15}{1.36}[0.6sin(t)-cos(t)]+100 and this will be your range of temperatures.