Math

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} $.

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 value of tan(135°) using the unit circle

To find the value of $ \tan(135^\circ) $ using the unit circle, we need to recall that $ \tan\theta $ is the ratio of the y-coordinate to the x-coordinate of the point where the terminal side of the angle intersects the unit circle.

The angle $ 135^\circ $ is in the second quadrant, where the tangent is negative. It corresponds to the reference angle $ 45^\circ $.

For $ 45^\circ $, the coordinates on the unit circle are:

$$ ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ) $$

In the second quadrant, the x-coordinate is negative, so the point is:

$$ (- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ) $$

Thus,

$$ \tan(135^\circ) = \frac{\frac{\sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} = -1 $$

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

Solve for the angle θ in the unit circle where sin(θ)cos(θ) = 1/4 and 0 ≤ θ < 2π

Given:
$$ \sin(\theta)\cos(\theta) = \frac{1}{4} $$

Using the double-angle identity:
$$ \sin(2\theta) = 2\sin(\theta)\cos(\theta) $$
We have:
$$ \sin(2\theta) = 2 \times \frac{1}{4} = \frac{1}{2} $$

Thus:
$$ 2\theta = \sin^{-1}(\frac{1}{2}) $$
Giving:
$$ 2\theta = \frac{\pi}{6} \text{ or } \frac{5\pi}{6} $$

Hence:
$$ \theta = \frac{\pi}{12} \text{ or } \frac{5\pi}{12} $$

Checking the interval $ 0 \leq \theta < 2\pi $:
The possible solutions are:
$$ \theta = \frac{\pi}{12}, \frac{5\pi}{12} \text{ or } \frac{13\pi}{12}, \frac{17\pi}{12} $$

Find the value of sin(2x) and cos(2x) on the unit circle

To find the value of $\sin(2x)$ and $\cos(2x)$ on the unit circle, we can utilize the double-angle formulas:

$$ \sin(2x) = 2\sin(x)\cos(x) $$

$$ \cos(2x) = \cos^2(x) – \sin^2(x) $$

Given a point on the unit circle (a, b) where $a = \cos(x)$ and $b = \sin(x)$, we can substitute:

$$ \sin(2x) = 2ab $$

$$ \cos(2x) = a^2 – b^2 $$