Unit Circle

Find the equation of the unit circle

The unit circle is defined as the set of all points in the coordinate plane that are exactly one unit away from the origin. The equation of the unit circle can be derived using the Pythagorean theorem. For a point $(x, y)$ on the circle:

$$ x^2 + y^2 = 1 $$

This equation represents all the points $(x, y)$ that satisfy the condition of being one unit away from the origin.

Identify the coordinates of the point on the unit circle at angle 7π/6

To find the coordinates of the point on the unit circle at angle $ \frac{7\pi}{6} $, use the unit circle values:

$$ \frac{7\pi}{6} $$

is in the third quadrant, where both sine and cosine are negative. The reference angle is $ \frac{\pi}{6} $, which corresponds to the coordinates:

$$ (\cos(\pi/6), \sin(\pi/6)) = (\frac{\sqrt{3}}{2}, \frac{1}{2}) $$

Since it is in the third quadrant, the coordinates are:

$$ \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) $$

The final coordinates are:

$$ \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) $$

Determine the angle θ in degrees for which the point (cos(θ), sin(θ)) is closest to the point (1/2, -sqrt(3)/2) on the unit circle

To find θ in degrees, we first find the angle whose coordinates on the unit circle are closest to (1/2, -√3/2). This point corresponds to the angle -60 degrees or 300 degrees.

The point (cos(θ), sin(θ)) that is closest must satisfy the equation:

$$ \cos(\theta) = \frac{1}{2} \text{ and } \sin(\theta) = -\frac{\sqrt{3}}{2} $$

Thus, the angle θ is:

$$ \theta = 300° $$

Find the secant of the angle when the point on the unit circle is at (sqrt(3)/2, 1/2)

Given the point $ \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) $ on the unit circle, we need to find the secant of the corresponding angle $ \theta $. Recall that $ \sec(\theta) = \frac{1}{\cos(\theta)} $ and $ \cos(\theta) $ is the x-coordinate.

So, $ \cos(\theta) = \frac{\sqrt{3}}{2} $. Hence,

$$ \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} $$

Determine the coordinates of a point on the unit circle with an angle of π/4

The unit circle is a circle with a radius of 1 centered at the origin (0, 0).

The coordinates of a point on the unit circle with an angle $ \frac{\pi}{4} $ are found using trigonometric functions:

$$ x = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

$$ y = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

Therefore, the coordinates are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

Find the exact values of sin(x), cos(x), and tan(x) for x = 7π/6 using the unit circle

To find the exact values of $ \sin(x) $, $ \cos(x) $, and $ \tan(x) $ for $ x = \frac{7\pi}{6} $, follow these steps:

The angle $ \frac{7\pi}{6} $ is in the third quadrant.

For the sine function:

$$ \sin\left(\frac{7\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2} $$

For the cosine function:

$$ \cos\left(\frac{7\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} $$

For the tangent function:

$$ \tan\left(\frac{7\pi}{6}\right) = \frac{\sin\left(\frac{7\pi}{6}\right)}{\cos\left(\frac{7\pi}{6}\right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

Find the sine and cosine values at the angle pi/4

At the angle $ \frac{\pi}{4} $, the coordinates on the unit circle are:

$$ \left( \cos\left( \frac{\pi}{4} \right), \sin\left( \frac{\pi}{4} \right) \right) $$

Using the unit circle values:

$$ \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}, \;\sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

Determine the quadrant of a given angle in radians on the unit circle

To determine the quadrant of an angle $ \theta $ in radians on the unit circle, follow these steps:

1. If $ \theta $ is greater than $ 2\pi $ or less than $ -2\pi $, reduce it by subtracting or adding multiples of $ 2\pi $ until it is within the range $ [0, 2\pi] $.

2. Check the reduced angle:

– If $ 0 \leq \theta < \frac{\pi}{2} $, the angle is in Quadrant I.

– If $ \frac{\pi}{2} \leq \theta < \pi $, the angle is in Quadrant II.

– If $ \pi \leq \theta < \frac{3\pi}{2} $, the angle is in Quadrant III.

– If $ \frac{3\pi}{2} \leq \theta < 2\pi $, the angle is in Quadrant IV.

Determine the quadrant of an angle of 45 degrees

To determine the quadrant of an angle of $45^{\circ}$, we need to look at the unit circle. Angles are measured counterclockwise from the positive x-axis.

Since $45^{\circ}$ is between $0^{\circ}$ and $90^{\circ}$, it lies in the first quadrant.

Therefore, the angle $45^{\circ}$ lies in the first quadrant.

Find the probability density function (pdf) for a uniform distribution on the unit circle

To find the probability density function (pdf) for a uniform distribution on the unit circle, we start by noting that the unit circle can be expressed in terms of its angular coordinate $\theta$, where $0 \leq \theta < 2\pi$.

Since the distribution is uniform, the probability density function (pdf) must be constant. The integral of the pdf over the entire circle must be 1:

$$ \int_0^{2\pi} f(\theta) \, d\theta = 1 $$

Let $f(\theta) = c$ be the constant pdf. Then:

$$ c \int_0^{2\pi} \, d\theta = 1 $$

Evaluating the integral gives:

$$ c \cdot 2\pi = 1 $$

Solving for $c$, we get:

$$ c = \frac{1}{2\pi} $$

Therefore, the pdf for a uniform distribution on the unit circle is:

$$ f(\theta) = \frac{1}{2\pi}, \quad 0 \leq \theta < 2\pi $$