Problem 9 (#16)
In order to figure out when the amount of sunlight becomes 12 hours for the first time, we need to have [tex]S(t)=12[/tex] and solve the function for [tex]t[/tex]:
[tex]\displaystyle S(t)=12+2\sin\biggr(\frac{2\pi}{365}t\biggr)\\\\12=12+2\sin\biggr(\frac{2\pi}{365}t\biggr)\\\\0=2\sin\biggr(\frac{2\pi}{365}t\biggr)\\ \\0=\sin\biggr(\frac{2\pi}{365}t\biggr)\\\\\pi=\frac{2\pi}{365}t\\\\365\pi=2\pi t\\\\\frac{365}{2}=t[/tex]
Thus, B is the correct answer
Problem 10 (#17)
- Recall the half-angle identity [tex]\displaystyle \tan\frac{x}{2}=\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}[/tex]
- Hence, [tex]\displaystyle \cot\frac{1}{2}x=\frac{1}{\tan\frac{1}{2}x }=\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}[/tex]
So, you could techincally say that the first two options work as each function has their respective identities true for each option.
Thus, I and II is the correct answer
Problem 11 (#18)
[tex]f(\theta)=g(\theta)\\\\4\sin\theta+1=\cos2\theta\\\\4\sin\theta+1=1-2\sin^2x\\\\2\sin^2\theta+4\sin\theta=0\\\\2\sin\theta\bigr(\sin\theta+2)=0[/tex]
[tex]\displaystyle 2\sin\theta=0\\\\\sin\theta=0\\\\\theta=\{0,\pi\}[/tex]
[tex]\sin\theta+2=0\\\\\sin\theta=-2[/tex]
The solution is indeterminate since -2 does not fall in the range of [tex][-1,1][/tex].
Thus, A is the correct answer
Problem 12 (#19)
Recall that the period of a tangent function [tex]y=a\tan(bx-c)+d[/tex] is [tex]\frac{\pi}{|b|}[/tex]. Hence, if [tex]b=2[/tex], then the period of the tangent function is [tex]\frac{\pi}{2}[/tex]. Since I can't see the graphs, you need to identify which graph has a period of [tex]\frac{\pi}{2}[/tex] (meaning the distance between two vertical asymptotes is pi/2), or post this problem again with the graphs.
