consider the following.\n$x=sin (\frac{1}{2}\theta )$, $y=cos (\frac{1}{2}\theta )$, $-pi leq \theta leq…

Explanation:

Step1: Square both equations

Given (x = \sin(\frac{1}{2}\theta)) and (y=\cos(\frac{1}{2}\theta)). Square (x): (x^{2}=\sin^{2}(\frac{1}{2}\theta)) Square (y): (y^{2}=\cos^{2}(\frac{1}{2}\theta))

Step2: Use the trigonometric identity

We know the Pythagorean identity (\sin^{2}\alpha+\cos^{2}\alpha = 1). Here (\alpha=\frac{1}{2}\theta). Add the two squared - equations: (x^{2}+y^{2}=\sin^{2}(\frac{1}{2}\theta)+\cos^{2}(\frac{1}{2}\theta))

Step3: Simplify the equation

Since (\sin^{2}(\frac{1}{2}\theta)+\cos^{2}(\frac{1}{2}\theta)=1), the Cartesian equation is (x^{2}+y^{2}=1). Now, consider the range of the parameter (\theta\in[-\pi,\pi]). When (\theta =-\pi), (x=\sin(-\frac{\pi}{2})=- 1), (y=\cos(-\frac{\pi}{2}) = 0) When (\theta=\pi), (x=\sin(\frac{\pi}{2}) = 1), (y=\cos(\frac{\pi}{2})=0)

For the direction: As (\theta) increases from (-\pi) to (\pi), we can also use another approach. Let (t=\frac{1}{2}\theta), then (\theta = 2t) and (t\in[-\frac{\pi}{2},\frac{\pi}{2}]) (x=\sin t), (y = \cos t). As (t) (or (\theta)) increases, for (x=\sin t), when (t) increases from (-\frac{\pi}{2}) to (\frac{\pi}{2}), (x) increases from (-1) to (1)

Answer:

(a) The Cartesian equation is (x^{2}+y^{2}=1). (b) The curve is the lower - half of the unit circle (x^{2}+y^{2}=1) (since when (\theta\in[-\pi,\pi]), (y = \cos(\frac{1}{2}\theta)\geq0) when (\theta\in[-\pi,\pi]) is not correct. Actually, when (\theta\in[-\pi,\pi]), (y=\cos(\frac{1}{2}\theta)) takes values from (0) (when (\theta=\pm\pi)) to (1) (when (\theta = 0)). The direction is from ((-1,0)) to ((1,0)) along the lower - half of the unit circle (x^{2}+y^{2}=1) (the first two graphs in the given options, the one with the arrow from (-1) on the (x) - axis to (1) on the (x) - axis along the lower - half of the circle is correct).