Math

Given a point P on the unit circle such that its coordinates are (cos(θ), sin(θ)), find the coordinates of the point Q, which is the reflection of P across the line y = x Then, find the coordinates of the point R, which is the reflection of Q across the

To find the coordinates of the point $Q$, which is the reflection of $P$ across the line $y = x$, we switch the coordinates of $P$. Therefore, the coordinates of $Q$ are $(sin(\theta), cos(\theta))$.

Next, to find the coordinates of the point $R$, which is the reflection of $Q$ across the $x$-axis, we negate the y-coordinate of $Q$. Thus, the coordinates of $R$ are $(sin(\theta), -cos(\theta))$.

Summary:
Coordinates of $Q$: $(sin(\theta), cos(\theta))$
Coordinates of $R$: $(sin(\theta), -cos(\theta))$

What is the cosine and sine of the angle π/4 on the unit circle?

To find the cosine and sine of the angle $\frac{\pi}{4}$ on the unit circle, we need to recall the coordinates of the point where the terminal side of the angle intersects the unit circle.

For the angle $\frac{\pi}{4}$, both the x-coordinate (cosine) and y-coordinate (sine) are equal. Since the unit circle has a radius of 1, we use the fact that $\cos(\theta) = \sin(\theta) = \frac{\sqrt{2}}{2}$ for this specific angle. Therefore,

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

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

Find the cotangent of \( \frac{\pi}{4} \) on the unit circle

To find the cotangent of $ \frac{\pi}{4} $ on the unit circle, we use the definition of cotangent in terms of sine and cosine.

$$ \cot \theta = \frac{\cos \theta}{\sin \theta} $$

For $ \theta = \frac{\pi}{4} $, both $ \sin \frac{\pi}{4} $ and $ \cos \frac{\pi}{4} $ are $ \frac{\sqrt{2}}{2} $.

Therefore,

$$ \cot \frac{\pi}{4} = \frac{\cos \frac{\pi}{4}}{\sin \frac{\pi}{4}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

The cotangent of $ \frac{\pi}{4} $ is 1.

Find the value of sec(θ) using the unit circle when θ = 2π/3, and verify the result using three different methods

First, we find the coordinates of the point on the unit circle corresponding to $\theta = \frac{2\pi}{3}$.

The coordinates are $\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)$.

Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, we have:

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

Verification using the Pythagorean identity:

$$\sec^2(\theta) = 1 + \tan^2(\theta)$$

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

$$\sec^2\left(\frac{2\pi}{3}\right) = 1 + 3 = 4$$

$$\sec\left(\frac{2\pi}{3}\right) = \pm 2 = -2.$$

Determine the cosine value at specific points on the unit circle

$$\text{Consider the point where the angle is } 60^\circ \text{ on the unit circle.}$$

$$\text{The cosine of } 60^\circ \text{ is given by } \cos(60^\circ) = \frac{1}{2}. $$

$$\text{Therefore, the cosine value at } 60^\circ \text{ on the unit circle is } \frac{1}{2}. $$

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

First, locate the angle $\frac{5\pi}{6}$ on the unit circle.

The angle $\frac{5\pi}{6}$ is in the second quadrant.

In the second quadrant, sine is positive and cosine is negative.

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

From the unit circle, $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

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

Find the value of tan for given angles on the unit circle

Consider the angle $\theta = \frac{3\pi}{4}$ on the unit circle.

First, determine the reference angle. The reference angle for $\frac{3\pi}{4}$ is $\frac{\pi}{4}$.

Since $\frac{3\pi}{4}$ is in the second quadrant, tangent is negative.

We know $\tan \frac{\pi}{4} = 1$, so:

$$\tan \frac{3\pi}{4} = -\tan \frac{\pi}{4} = -1$$

Cosine Values on the Unit Circle

Consider the point $P(\frac{1}{2}, \frac{\sqrt{3}}{2})$ on the unit circle. Determine the cosine of the angle $\theta$ corresponding to this point.

Solution:

On the unit circle, the coordinates of a point $P(x, y)$ correspond to $(\cos \theta, \sin \theta)$. Given the coordinates $P(\frac{1}{2}, \frac{\sqrt{3}}{2})$, we can identify that $\cos \theta = \frac{1}{2}$.

Thus, the cosine of the angle $\theta$ is:

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

Find the slope of the tangent line to the unit circle at the point where $\theta = \frac{\pi}{4}$

To find the slope of the tangent line to the unit circle at the point where $\theta = \frac{\pi}{4}$, we start by finding the coordinates of the point on the unit circle.

At $\theta = \frac{\pi}{4}$, the coordinates are:

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

The slope of the tangent line to the unit circle at any point $(x, y)$ is given by $-\frac{x}{y}$.

Therefore, the slope at the point $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ is:

$$ -\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$