Unit Circle

Find the coordinates of the point on the unit circle for angle π/3

For the angle $ \frac{\pi}{3} $ on the unit circle, the coordinates are found using the sine and cosine functions.

The x-coordinate is:

$$ \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} $$

The y-coordinate is:

$$ \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $$

Thus, the coordinates are:

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

Determine the value of cos(7π/6) using the unit circle

To determine the value of $ \cos\left(\frac{7\pi}{6}\right) $ using the unit circle, we need to locate the angle $ \frac{7\pi}{6} $ in radians. This angle is in the third quadrant.

In the third quadrant, the cosine function is negative. The reference angle for $ \frac{7\pi}{6} $ is $ \frac{\pi}{6} $, whose cosine value is $ \frac{\sqrt{3}}{2} $.

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

Determine the sine and cosine values for an angle of 5π/6 radians on the unit circle

To find the sine and cosine of $ \frac{5\pi}{6} $ on the unit circle, we use the reference angle and the fact that it lies in Quadrant II:

The reference angle for $ \frac{5\pi}{6} $ is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.

In Quadrant II, sine is positive and cosine is negative. Therefore:

$$ \sin\left( \frac{5\pi}{6} \right) = \sin\left( \frac{\pi}{6} \right) = \frac{1}{2} $$

$$ \cos\left( \frac{5\pi}{6} \right) = -\cos\left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2} $$

Express the coordinates of key points on the unit circle in terms of trigonometric functions

To express the coordinates of key points on the unit circle in terms of trigonometric functions, remember that each point on the unit circle corresponds to an angle $\theta$ and can be written as $(\cos(\theta), \sin(\theta))$. For example:

For $\theta = 0$: $$\cos(0) = 1, \sin(0) = 0$$ Hence, the coordinates are $(1, 0)$.

For $\theta = \frac{\pi}{2}$: $$\cos\left(\frac{\pi}{2}\right) = 0, \sin\left(\frac{\pi}{2}\right) = 1$$ Hence, the coordinates are $(0, 1)$.

For $\theta = \pi$: $$\cos(\pi) = -1, \sin(\pi) = 0$$ Hence, the coordinates are $(-1, 0)$.

For $\theta = \frac{3\pi}{2}$: $$\cos\left(\frac{3\pi}{2}\right) = 0, \sin\left(\frac{3\pi}{2}\right) = -1$$ Hence, the coordinates are $(0, -1)$.

Find the value of arcsin(x) for x = sqrt(3)/2 on the unit circle

To find the value of $ \arcsin(x) $ for $ x = \sqrt{3}/2 $ on the unit circle, we need to determine the angle $ \theta $ such that $ \sin(\theta) = \sqrt{3}/2 $ and $ \theta $ lies in the range $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.

The angle $ \theta $ corresponding to $ \sin(\theta) = \sqrt{3}/2 $ is $ \frac{\pi}{3} $.

Hence, $ \arcsin(\sqrt{3}/2) = \frac{\pi}{3} $.

Find the sine and cosine values at different angles on the unit circle

Given the unit circle, find the sine and cosine values for the following angles:

1. $0$ radians

2. $\frac{\pi}{4}$ radians

3. $\frac{\pi}{2}$ radians

1. At $0$ radians, the coordinates are $(1, 0)$, so the sine value is $0$ and the cosine value is $1$.

2. At $\frac{\pi}{4}$ radians, the coordinates are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$, so the sine value is $\frac{\sqrt{2}}{2}$ and the cosine value is $\frac{\sqrt{2}}{2}$.

3. At $\frac{\pi}{2}$ radians, the coordinates are $(0, 1)$, so the sine value is $1$ and the cosine value is $0$.

Find the angle in radians for the point (-1/2, -√3/2) on the unit circle

To find the angle in radians for the point $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$ on the unit circle, we first identify the coordinates. These coordinates correspond to one of the 30-60-90 triangle

Determine the coordinates of a point on the unit circle with a given angle

To determine the coordinates of a point on the unit circle given the angle $ \theta $, use the unit circle formulas:

$$ x = \cos(\theta) $$

$$ y = \sin(\theta) $$

For example, if $ \theta = 60^\circ $:

$$ x = \cos(60^\circ) = \frac{1}{2} $$

$$ y = \sin(60^\circ) = \frac{\sqrt{3}}{2} $$

So the coordinates are $ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $.

Find the length of the arc subtended by a central angle of θ radians in a unit circle

To find the length of the arc subtended by a central angle $ \theta $ radians in a unit circle, we use the formula:

$$ s = r \theta $$

Here, the radius $ r $ of a unit circle is 1. So:

$$ s = 1 \cdot \theta $$

Therefore, the length of the arc is:

$$ s = \theta $$

Find the exact value of tan(θ) given that sin(θ) = 3/5 and θ is in the second quadrant

Given that $ \sin(\theta) = \frac{3}{5} $ and $ \theta $ is in the second quadrant:

Since $ \sin(\theta) $ is positive in the second quadrant, $ \cos(\theta) $ must be negative:

Use the Pythagorean identity:

$$ \sin^2(\theta) + \cos^2(\theta) = 1 $$

Substitute $ \sin(\theta) = \frac{3}{5} $:

$$ \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 $$

$$ \frac{9}{25} + \cos^2(\theta) = 1 $$

$$ \cos^2(\theta) = 1 – \frac{9}{25} = \frac{16}{25} $$

Since $ \theta $ is in the second quadrant, $ \cos(\theta) $ is negative:

$$ \cos(\theta) = -\frac{4}{5} $$

Now find $ \tan(\theta) $:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} $$

Thus, $ \tan(\theta) = -\frac{3}{4} $.