Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 04/Solution 2

(Alternative solution) By Hint 2, we know that $\sin \left({\frac {2\pi t}{L}}\right)$ and $\cos \left({\frac {2\pi t}{L}}\right)$ are $L$-periodic.

This implies that for any constant $a$ and $b\neq 0$, both functions $a+b\sin \left({\frac {2\pi t}{L}}\right)$ and $a+b\cos \left({\frac {2\pi t}{L}}\right)$ are also $L$-periodic;

Indeed, putting $T(t)=a+b\sin \left({\frac {2\pi t}{L}}\right)$, we have $T(t+L)=a+b\sin \left({\frac {2\pi (t+L)}{L}}\right)=a+b\sin \left({\frac {2\pi t}{L}}\right)=T(t)$.

Similar argument holds for $a+b\cos \left({\frac {2\pi t}{L}}\right)$.

Therefore, (i), (ii), (iii) are $12$-periodic, while (iv), (v), (vi) are $24$-periodic.

However, since at $t=6$ (6 am) and at $t=18=6+12$ (6 pm), the function values of $T(t)$ are different, i.e., $T(6)=37.6\neq 36.8=T(18)$,

$T(t)$ is not $12$-periodic---(i), (ii), (iii) are out.

Then, using $\cos \left({\frac {2\pi \cdot 6}{24}}\right)=\cos \left({\frac {\pi }{2}}\right)=0$, $T$ given in (v) and (vi) satisfies $T(6)=37.2$, which is different from the given information $T(6)=37.6$---(v), (vi) are out.

Answer $\color {blue}(iv)$