Respuesta :
First solve for theta in 'x' equation:
[tex]\theta = 2 sin^{-1} x[/tex]
Substitute into 'y' equation:
[tex]y = cos (\frac{2 sin^{-1} x}{2}) = cos (sin^{-1} x)[/tex]
Simplify by rewriting cos in terms of sine.
Using pythagorean identity: [tex]sin^2 + cos^2 = 1[/tex]
[tex]cos = \sqrt{1-sin^2} \\ \\ y = \sqrt{1-sin^2 (sin^{-1} x)} [/tex]
Simplify using inverse function property: [tex]f(f^{-1} (x)) = x[/tex]
[tex]y = \sqrt{1 - x^2}[/tex]
[tex]\theta = 2 sin^{-1} x[/tex]
Substitute into 'y' equation:
[tex]y = cos (\frac{2 sin^{-1} x}{2}) = cos (sin^{-1} x)[/tex]
Simplify by rewriting cos in terms of sine.
Using pythagorean identity: [tex]sin^2 + cos^2 = 1[/tex]
[tex]cos = \sqrt{1-sin^2} \\ \\ y = \sqrt{1-sin^2 (sin^{-1} x)} [/tex]
Simplify using inverse function property: [tex]f(f^{-1} (x)) = x[/tex]
[tex]y = \sqrt{1 - x^2}[/tex]
Answer:
[tex]x^{2}+y^{2}=1[/tex] Standard form.
Step-by-step explanation:
Parametric equations are useful to describe the motion of particles for instance, among other applications. In a trigonometric circle we can find its coordinates.
1) Rewriting for the sake of clarity:
[tex]x=sin\frac{1}{2}\theta \:and\:\: y=cos\frac{1}{2}\theta \:,-\pi\leq \theta \leq \pi[/tex]
2) Solving, and using the Pythagorean Identity:
[tex]x=sen\frac{1}{2}\theta ,y=cos\frac{1}{2}\theta \Rightarrow sen^{2}x+cos^{2}y=1\Rightarrow \left ( sen\frac{1}{2}\theta\right)^{2}+\left ( cos\frac{1}{2}\theta \right )^{2}=1\\\:Replacing\:the\:parameter\Rightarrow x^2+y^2=1[/tex]
Since the value for [tex]\theta[/tex] angle varies from [tex]-\pi\leqslant\theta \leqslant \pi[/tex] then this is a description of a half of a circumference.
Check the graph below.
