Answer:
[tex]y = - \frac{ \sqrt{3} }{3}x [/tex]
[tex] {x}^{2} + {y}^{2} = 1[/tex]
Step-by-step explanation:
You are correct with the graph, the angle
[tex] \frac{5\pi}{6} [/tex]
should lies in the second quadrant so the polar graph will lie in the second quadrant then go to the fourth quadrant. It would be a line not a circle.
For the equation, use the tangent equation because it has y and x in it
[tex] \alpha = \frac{5\pi}{6} [/tex]
[tex] \tan( \alpha ) = \tan( \frac{5\pi}{6} ) [/tex]
[tex] \frac{y}{x} = - \frac{ \sqrt{3} }{3} [/tex]
[tex]y = - \frac{ \sqrt{3} }{3} x[/tex]
For the next picture, the graph is correct,
We know that
[tex] \cos {}^{2} (3t) + \sin {}^{2} ( {}^{}3t ) = 1[/tex]
[tex] {x}^{2} + {y}^{2} = 1[/tex]
Alternate Way:
[tex]x = \cos(3t) [/tex]
[tex] \frac{ \cos {}^{ - 1} (x) }{3} = t[/tex]
[tex]y = \sin(3( \frac{ \cos {}^{ - 1} (x) }{3} ) [/tex]
[tex]y = \sin( \cos {}^{ - 1} (x) ) [/tex]
Here we must imagine a triangle such that an angle has a hypotenuse 1, and a leg x, by definition the missing side is
[tex] \sqrt{1 - {x}^{2} } [/tex]
by using the Pythagorean theorem.
[tex] \cos( \alpha ) = \frac{x}{1} [/tex]
so
[tex] \sin( \alpha ) = \frac{1 - \sqrt{ {x}^{2} } }{1} [/tex]
[tex]y = \sqrt{1 - {x}^{2} } [/tex]
[tex] {y}^{2} = 1 - {x}^{2} [/tex]
[tex] {x}^{2} + {y }^{2} = 1[/tex]
