x = sin 1/2θ, y = cos 1/2θ, - π ≤ θ ≤ π(a) eliminate the parameter to find a cartesian equation of the curve.

Solution:

Given y = cos 1/2θ implies that the radius of the circle is 1 because the general equation would be y = r cos 1/2θ.

Therefore in this case r = 1.

Since x = sin 1/2θ, it also implies that r = 1 because the general form of the equation would be x = r sin 1/2θ.

Squaring x and y we get the following equation:

x² = sin²1/2θ -------(1)

y² = cos² 1/2θ ---------(2)

Adding 1 and 2 we have

x² + y² = cos² 1/2θ + sin² 1/2θ ----- (3)

The RHS of equation (3) is the fundamental trigonometric identity which is equal to 1 i.e.

cos² 1/2θ + sin²1/2θ = 1

Hence equation (3) can be written as

x² + y² = 1

⇒ y = \( \sqrt{1-x^{2}} \) -1 ≤ x ≤ 1

Hence we obtain the cartesian curve from the equation in polar form.

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x = sin 1/2θ, y = cos 1/2θ, - π ≤ θ ≤ π(a) eliminate the parameter to find a cartesian equation of the curve.

Summary:

The cartesian equation for the above polar equation form is: y = \( \sqrt{1-x^{2}} \) -1 ≤ x ≤ 1