7. trigonometric functions\na. identifying trigonometric functions which equation represents the graph…

Explanation:

Step1: Check option i

For $f(\theta)=\sin(\theta)-2$, when $\theta = \frac{\pi}{2}$, $f(\frac{\pi}{2})=\sin(\frac{\pi}{2})-2=1 - 2=-1$, but the graph has $f(\frac{\pi}{2}) = 1$, so option i is incorrect.

Step2: Check option ii

For $f(\theta)=\cos(\theta+\pi)=-\cos(\theta)$. When $\theta=\frac{\pi}{2}$, $f(\frac{\pi}{2})=-\cos(\frac{\pi}{2}) = 0$, but the graph has $f(\frac{\pi}{2})=1$, so option ii is incorrect.

Step3: Check option iii

For $f(\theta)=2\sin(\theta)-1$. When $\theta=\frac{\pi}{2}$, $f(\frac{\pi}{2})=2\sin(\frac{\pi}{2})-1=2\times1 - 1 = 1$. When $\theta=\frac{3\pi}{2}$, $f(\frac{3\pi}{2})=2\sin(\frac{3\pi}{2})-1=2\times(- 1)-1=-3$. When $\theta=\frac{\pi}{6}$, $f(\frac{\pi}{6})=2\sin(\frac{\pi}{6})-1=2\times\frac{1}{2}-1 = 0$. When $\theta=\frac{5\pi}{6}$, $f(\frac{5\pi}{6})=2\sin(\frac{5\pi}{6})-1=2\times\frac{1}{2}-1 = 0$. When $\theta=\frac{13\pi}{6}$, $f(\frac{13\pi}{6})=2\sin(\frac{13\pi}{6})-1=2\times\frac{1}{2}-1 = 0$. This function matches the points on the graph.

Step4: Check option iv

For $f(\theta)=2\cos(\theta - 1)$, when $\theta=\frac{\pi}{2}$, $f(\frac{\pi}{2})=2\cos(\frac{\pi}{2}-1)\neq1$, so option iv is incorrect.

Answer:

iii. $f(\theta)=2\sin(\theta)-1$