Science:Math Exam Resources/Courses/MATH102/December 2014/Question B 04/Solution 1

The general expression of a cosine function is $T(t)=A+B\cos (\omega t+\phi ).$ Suppose $A$, $B$ are both positive. Here the lowest temperature is $2^o$ C and highest is $10^o$ C, then

$A-B=2,A+B=10,$

so $A=6$ and $B=4$. The value $2\pi /\omega$ is the period of this function. As it describes the temperature changes throughout a day which lasts for 24 hours, we have $\frac{2\pi }{\omega }=24,\omega =\frac{2\pi }{24}.$ From the problem statement we know the lowest temperature is obtained at 7:00, by symmetry the highest is around 19:00. Then to shift the maxima from $0$ to $19$ we subtract $t$ by 19. So $\phi =-19\frac{2\pi }{24}$ and the final function is $T(t)=6+4\cos (\frac{\pi }{12}t-\frac{\pi }{12}19).$

Note the expression of the solution is not unique, it can be written in infinitely many ways, such as $T(t)=6+4\cos (\frac{\pi }{12}t+\frac{\pi }{12}5),$