Math

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} $.

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 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} $.

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

To find the exact values of $\sin(\frac{7\pi}{6})$, $\cos(\frac{7\pi}{6})$, and $\tan(\frac{7\pi}{6})$ using the unit circle, we follow these steps:

1. Identify the reference angle: The reference angle for $\frac{7\pi}{6}$ is $\frac{\pi}{6}$.

2. Determine the quadrant: Since $\frac{7\pi}{6}$ is in the third quadrant, both sine and cosine are negative.

3. Evaluate sine and cosine: $$ \sin(\frac{\pi}{6}) = \frac{1}{2}, \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $$

Thus, $$ \sin(\frac{7\pi}{6}) = -\frac{1}{2}, \cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2} $$

4. Compute tangent: $$ \tan(\frac{7\pi}{6}) = \frac{\sin(\frac{7\pi}{6})}{\cos(\frac{7\pi}{6})} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

Thus, the exact values are: $$ \sin(\frac{7\pi}{6}) = -\frac{1}{2}, \cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2}, \tan(\frac{7\pi}{6}) = \frac{\sqrt{3}}{3} $$

Find the values of sin, cos, and tan for 30 degrees on the unit circle

For $ 30^\circ $ on the unit circle:

$$ \sin(30^\circ) = \frac{1}{2} $$

$$ \cos(30^\circ) = \frac{\sqrt{3}}{2} $$

$$ \tan(30^\circ) = \frac{1}{\sqrt{3}} \text{ or } \frac{\sqrt{3}}{3} $$

Find the coordinates of the vertices of a triangle inscribed in a unit circle given angles

Given the angles $ \theta_1, \theta_2, \theta_3 $ of the vertices of the triangle, the coordinates of the vertices on the unit circle are:

Vertex 1: $ ( \cos(\theta_1), \sin(\theta_1) ) $

Vertex 2: $ ( \cos(\theta_2), \sin(\theta_2) ) $

Vertex 3: $ ( \cos(\theta_3), \sin(\theta_3) ) $

Let