Unit Circle

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

Find the values of sin, cos, and tan for an angle of π/4 on the unit circle

To find the values of $ \sin, \cos, $ and $ \tan $ for an angle of $ \frac{\pi}{4} $ on the unit circle, we start with:

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

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

$$ \tan\left(\frac{\pi}{4}\right) = 1 $$

Find the value of arctan(sin(3π/4))

To find the value of $ \arctan(\sin(\frac{3\pi}{4})) $, we first need to find the value of $ \sin(\frac{3\pi}{4}) $.

$$ \sin(\frac{3\pi}{4}) = \sin(\pi – \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$

Now, we need to determine the value of $ \arctan(\frac{\sqrt{2}}{2}) $.

Since $ \arctan(x) $ is the inverse of $ \tan(x) $, we seek an angle $ \theta $ such that:

$$ \tan(\theta) = \frac{\sqrt{2}}{2} $$

One such angle is $ \theta = \frac{\pi}{4} $, but considering the range of $ \arctan $, the solution is:

$$ \arctan(\sin(\frac{3\pi}{4})) = \arctan(\frac{\sqrt{2}}{2}) = \frac{\pi}{4} $$

Determine the values of sin(θ), cos(θ), and tan(θ) for θ in the second quadrant of the unit circle

In the second quadrant, the angle $ \theta $ ranges from $ \frac{\pi}{2} $ to $ \pi $. Here, $ \sin(\theta) $ is positive, $ \cos(\theta) $ is negative, and $ \tan(\theta) $ is negative.

Using the unit circle, for $ \theta = \frac{2\pi}{3} $:

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

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

$$ \tan(\frac{2\pi}{3}) = -\sqrt{3} $$