Unit Circle

Find the sine and cosine values for the angle 5π/6

To find the sine and cosine values for the angle $\frac{5\pi}{6}$, we first understand that this angle is located in the second quadrant of the unit circle.

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

We know the sine and cosine values for $\frac{\pi}{6}$ are $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

Since $\frac{5\pi}{6}$ is in the second quadrant, the sine value remains positive, and the cosine value becomes negative.

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

Determine the coordinates of the point where the terminal side of an angle θ = 5π/4 radians intersects the unit circle

To find the coordinates of the point where the terminal side of an angle $\theta = \frac{5\pi}{4}$ intersects the unit circle, we start by expressing the angle in degrees:

$$\theta = \frac{5\pi}{4} \cdot \frac{180}{\pi} = 225^{\circ}$$

This angle is in the third quadrant where both sine and cosine are negative. For the unit circle, we can use the reference angle:

$$ 225^{\circ} – 180^{\circ} = 45^{\circ} $$

The coordinates corresponding to $45^{\circ}$ are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

Since $225^{\circ}$ is in the third quadrant:

$$ \cos(225^{\circ}) = -\frac{\sqrt{2}}{2}, \sin(225^{\circ}) = -\frac{\sqrt{2}}{2} $$

Therefore, the coordinates are:

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

Find the value of cos(θ) and sin(θ) for θ = 225°

First, we need to find the reference angle for $\theta = 225^\circ$. Since $225^\circ$ is in the third quadrant, the reference angle is:

$$225^\circ – 180^\circ = 45^\circ$$

In the third quadrant, the cosine and sine values are negative. For a $45^\circ$ reference angle, we have:

$$\cos(45^\circ) = \frac{\sqrt{2}}{2}$$ and $$\sin(45^\circ) = \frac{\sqrt{2}}{2}$$

Thus, in the third quadrant:

$$\cos(225^\circ) = -\frac{\sqrt{2}}{2}$$

$$\sin(225^\circ) = -\frac{\sqrt{2}}{2}$$

Find the value of cos(-π/3) using the unit circle

To find $\cos(-\pi / 3)$, we can start by recognizing that the cosine function is even. This means $\cos(-x) = \cos(x)$. Therefore:

$$\cos(-\pi / 3) = \cos(\pi / 3)$$

From the unit circle, we know that:

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

So, the value of $\cos(-\pi / 3)$ is:

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

Find the equation of the unit circle centered at the origin

To find the equation of the unit circle centered at the origin, we start with the standard form of the circle equation:

$$ (x – h)^2 + (y – k)^2 = r^2 $$

For a unit circle centered at the origin, the center (h, k) is (0, 0) and the radius r is 1. Substituting these values, we get:

$$ (x – 0)^2 + (y – 0)^2 = 1^2 $$

Simplifying this, the equation of the unit circle is:

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

Find the tangent of the angle

Given an angle \( \theta = \frac{\pi}{4} \), find \( \tan(\theta) \).

Since the angle \( \theta \) is within the first quadrant and \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we have:

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

Therefore:

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

So, \( \tan(\frac{\pi}{4}) = 1 \).

Given a point on the unit circle at an angle of 5π/4 radians, find the coordinates of this point Then, if the unit circle is flipped about the y-axis, determine the new coordinates of the original point after the flip

First, find the coordinates of the point on the unit circle at $\frac{5\pi}{4}$ radians. This point can be represented as:

$$ (\cos(\frac{5\pi}{4}), \sin(\frac{5\pi}{4})) $$

We know that:

$$ \cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} $$

$$ \sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} $$

Thus, the coordinates at $\frac{5\pi}{4}$ radians are:

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

Next, when the unit circle is flipped about the y-axis, the x-coordinate of the point changes sign, but the y-coordinate remains the same. Therefore, the new coordinates after the flip are:

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

Find the values of cotangent for a given angle on the unit circle and verify their consistency

To find the values of $\cot(\theta)$ for $\theta = \frac{3\pi}{4}$ on the unit circle, we start by identifying the coordinates of this angle on the unit circle.

The coordinates for $\theta = \frac{3\pi}{4}$ are $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. The cotangent function is defined as the cosine divided by the sine of the angle: $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$.

Therefore,

$$\cot\left(\frac{3\pi}{4}\right) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1$$

The value of $\cot\left(\frac{3\pi}{4}\right)$ is -1.

What is the cosine of the angle at 2π/3 on the unit circle?

First, recognize that the angle $ \frac{2\pi}{3} $ is in the second quadrant.

In the unit circle, the cosine function is negative in the second quadrant.

The reference angle for $ \frac{2\pi}{3} $ is $ \pi – \frac{2\pi}{3} = \frac{\pi}{3} $.

We know that $ \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} $.

Therefore, the cosine of $ \frac{2\pi}{3} $ is:

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

Determine the Tangent Slope at a Given Point on the Unit Circle

Let the given point on the unit circle be $(a, b)$, where $a^2 + b^2 = 1$. We need to determine the slope of the tangent line at this point.

The equation of the unit circle is given by:

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

To find the slope of the tangent line at $(a, b)$, we first implicitly differentiate both sides of the equation with respect to $x$:

$$2x + 2y\frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = -\frac{x}{y}$$

Substituting the point $(a, b)$ into the derivative:

$$\frac{dy}{dx}\bigg|_{(a,b)} = -\frac{a}{b}$$

Therefore, the slope of the tangent line at the point $(a, b)$ is $-\frac{a}{b}$.