Unit Circle

Find the points on the unit circle where the secant of the angle is equal to 2, and prove their coordinates

To find points on the unit circle where $ \sec(\theta) = 2 $, recall that:

$$ \sec(\theta) = \frac{1}{\cos(\theta)} $$

Thus, we need:

$$ \frac{1}{\cos(\theta)} = 2 $$

So:

$$ \cos(\theta) = \frac{1}{2} $$

The angles on the unit circle with $ \cos(\theta) = \frac{1}{2} $ are:

$$ \theta = \frac{\pi}{3} \text{ and } \theta = \frac{5\pi}{3} $$

The corresponding points on the unit circle are:

For $ \theta = \frac{\pi}{3} $:

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

For $ \theta = \frac{5\pi}{3} $:

$$ (\cos(\frac{5\pi}{3}), \sin(\frac{5\pi}{3})) = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) $$

Determine the coordinates of the points where the unit circle intersects the line y = 2x + 1

First, recall the equation of the unit circle:

$$ x^2 + y^2 = 1 $$

Substitute $ y = 2x + 1 $ into the unit circle equation:

$$ x^2 + (2x + 1)^2 = 1 $$

Expand and simplify the equation:

$$ x^2 + 4x^2 + 4x + 1 = 1 $$

$$ 5x^2 + 4x = 0 $$

Factor the quadratic equation:

$$ x(5x + 4) = 0 $$

Thus, $ x = 0 $ or $ x = -\frac{4}{5} $. For $ x = 0 $:

$$ y = 2(0) + 1 = 1 $$

For $ x = -\frac{4}{5} $:

$$ y = 2\left(-\frac{4}{5}\right) + 1 = -\frac{8}{5} + 1 = -\frac{3}{5} $$

The intersection points are:

$$ (0, 1) \: \text{and} \: \left(-\frac{4}{5}, -\frac{3}{5}\right) $$

Find the angle in degrees for which the sine and cosine values are equal on the unit circle

To find the angle $\theta$ in degrees for which $\sin(\theta) = \cos(\theta)$ on the unit circle, start by equating the two trigonometric functions:

$$ \sin(\theta) = \cos(\theta) $$

Divide both sides by $\cos(\theta)$ (where $\cos(\theta) \neq 0$):

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

So, the tangent function is:

$$ \tan(\theta) = 1 $$

The angle $\theta$ for which $\tan(\theta) = 1$ is:

$$ \theta = 45^\circ $$

Find the coordinates of the point on the unit circle where the terminal side of the angle 5π/6 intersects the circle

To find the coordinates of the point where the terminal side of the angle $ \frac{5\pi}{6} $ intersects the unit circle, we use the unit circle definition:

The coordinates are given by:

$$ (\cos(\theta), \sin(\theta)) $$

For $ \theta = \frac{5\pi}{6} $:

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

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

Thus, the coordinates are:

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

Find the exact values of sin(θ), cos(θ), and tan(θ) for θ = 3π/4

Consider the angle $ \theta = \frac{3\pi}{4} $, which is in the second quadrant.

Using the unit circle, we know:

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

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

$$ \tan(\frac{3\pi}{4}) = \frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

Find the exact trigonometric values of cos(5π/6) and sin(5π/6) from the unit circle

To find the exact values of $\cos\left(\frac{5\pi}{6}\right)$ and $\sin\left(\frac{5\pi}{6}\right)$, we refer to the unit circle.

For the angle $\frac{5\pi}{6}$:

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

On the unit circle, the coordinates for $\frac{\pi}{6}$ are $(\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6}))$ = (\frac{\sqrt{3}}{2}, \frac{1}{2})$

Since $\frac{5\pi}{6}$ is in the second quadrant, $\cos(\frac{5\pi}{6})$ is negative and $\sin(\frac{5\pi}{6})$ is positive:

Thus, $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$ and $\sin(\frac{5\pi}{6}) = \frac{1}{2}$

Find the coordinates of a point on the unit circle at an angle of π/6

To find the coordinates of a point on the unit circle at an angle of $ \frac{\pi}{6} $, we use the fact that the coordinates are given by $ ( \cos(\theta), \sin(\theta)) $ where $ \theta $ is the angle:

$$ \theta = \frac{\pi}{6} $$

Therefore:

$$ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $$

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

The coordinates are:

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

Calculate the area of the shaded region in a unit circle with central angles

Let’s calculate the area of the shaded region in a unit circle with central angles $ \theta $ and $ \alpha $.

The area of a sector of a circle is given by:

$$ A = \frac{1}{2} r^2 \theta $$

For a unit circle, $ r = 1 $, so the above formula simplifies to:

$$ A = \frac{1}{2} \theta $$

The area of the shaded region is then the difference between two sector areas:

$$ A_{shaded} = \frac{1}{2} (\theta – \alpha) $$

Find the coordinates on the unit circle for the angles

Find the coordinates on the unit circle for the angle $ \theta = \frac{\pi}{4} $:

The coordinates are given by $ (\cos(\theta), \sin(\theta)) $.

For $ \theta = \frac{\pi}{4} $:

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

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

The coordinates are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

Find the tangent values at 0, π/4, and π/3 on the unit circle

To find the tangent values at points $0$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$ on the unit circle, we use the tangent function $tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$:

1. At $\theta = 0$:

$$ \tan(0) = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0 $$

2. At $\theta = \frac{\pi}{4}$:

$$ \tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

3. At $\theta = \frac{\pi}{3}$:

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