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

The general expression of a cosine function is $T(t)=A+B\cos(\omega t+\varphi ).$ 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 ${\frac {2\pi }{\omega }}=24,\quad \omega ={\frac {2\pi }{24}}.$ 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 $\varphi =-19{\frac {2\pi }{24}}$ and the final function is $T(t)=6+4\cos \left({\frac {\pi }{12}}t-{\frac {\pi }{12}}19\right).$

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