Math

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:

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