$ ext{Properties and Implications of the Unit Circle}$

$\text{A unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. To understand its properties, consider a point } (x, y) \text{ on the unit circle. According to the equation of a circle, we have:}$

$x^2 + y^2 = 1$

$\text{For example, if } x = \frac{1}{2}, \text{ then:}$

$\left( \frac{1}{2} \right)^2 + y^2 = 1$

$\frac{1}{4} + y^2 = 1$

$y^2 = 1 – \frac{1}{4}$

$y^2 = \frac{3}{4}$

$y = \pm \frac{\sqrt{3}}{2}$

Therefore, the point $( \frac{1}{2}, \pm \frac{\sqrt{3}}{2} )$ lies on the unit circle.

$ ext{The unit circle is defined by the equation}$

$x^2 + y^2 = 1$

$ ext{For a point } P(cos heta, sin heta) ext{ on the circle, its coordinates can be derived using trigonometric identities:}$

$cos^2 heta + sin^2 heta = 1$

$ ext{If } heta = 45^{circ}, ext{ then:}$

$cos 45^{circ} = sin 45^{circ} = frac{sqrt{2}}{2}$

$ ext{Thus, the coordinates at } heta = 45^{circ} ext{ are: } (frac{sqrt{2}}{2}, frac{sqrt{2}}{2})$

These points satisfy the unit circle equation.

$ ext{A unit circle has the equation:}$

$x^2 + y^2 = 1$

$ ext{For } x = frac{3}{4}, ext{ we get:}$

$left( frac{3}{4}
ight)^2 + y^2 = 1$

$frac{9}{16} + y^2 = 1$

$y^2 = frac{7}{16}$

$y = pm frac{sqrt{7}}{4}$

Therefore, $( frac{3}{4}, pm frac{sqrt{7}}{4} )$ lies on the unit circle.

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