Question

consider the following.
$x=sin (\frac{1}{2}\theta )$, $y=cos (\frac{1}{2}\theta )$, $-pi leq \theta leq pi$
(a) eliminate the parameter to find a cartesian equation of the curve.

(b) sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

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).