Unit Circle

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:

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

Find the exact values of sin, cos, and tan for the angle 225° using the unit circle

To find the exact values of $\sin$, $\cos$, and $\tan$ for the angle $225^{\circ}$ using the unit circle, we first note that $225^{\circ}$ is in the third quadrant.

In the third quadrant, both sine and cosine values are negative, and tangent value is positive since tangent is the ratio of sine to cosine.

The reference angle for $225^{\circ}$ is $225^{\circ} – 180^{\circ} = 45^{\circ}$.

The values for $45^{\circ}$ are:

$$ \sin 45^{\circ} = \frac{\sqrt{2}}{2} $$

$$ \cos 45^{\circ} = \frac{\sqrt{2}}{2} $$

Therefore, the values in the third quadrant (for $225^{\circ}$) are:

$$ \sin 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \cos 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \tan 225^{\circ} = \frac{\sin 225^{\circ}}{\cos 225^{\circ}} = \frac{- \frac{\sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} = 1 $$

Hence, the values are:

$$ \sin 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \cos 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \tan 225^{\circ} = 1 $$

Given that the point on the unit circle corresponding to the angle θ is (-3/5, -4/5), determine the value of θ in radians, ensuring that θ lies within the interval [0, 2π) Describe your method and calculations in detail

Given the point on the unit circle $\left( -\frac{3}{5}, -\frac{4}{5} \right)$, we need to determine the angle $\theta$ in radians.

First, note that the x and y coordinates tell us which quadrant the angle is in. Both coordinates are negative, so the point lies in the third quadrant.

The reference angle $\alpha$ can be determined using the tangent function:

$$ \tan \alpha = \left| \frac{y}{x} \right| = \left| \frac{-\frac{4}{5}}{-\frac{3}{5}} \right| = \frac{4}{3} $$

Using the arctangent function, we find:

$$ \alpha = \arctan \left( \frac{4}{3} \right) $$

Since this is a third quadrant angle, $\theta$ is given by:

$$ \theta = \pi + \alpha $$

Thus,

$$ \theta = \pi + \arctan \left( \frac{4}{3} \right) $$

Given a point P on the unit circle at an angle θ (in radians) from the positive x-axis, find the coordinates of P if θ is transformed by the function f(θ) = 2θ + π/4 Also, identify the new x and y coordinates after the transformation

Given the initial angle $ \theta $, the coordinates of the point $ P $ are:

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

With the transformation $ f(\theta) = 2\theta + \frac{\pi}{4} $, let the new angle be $ \theta’ = 2\theta + \frac{\pi}{4} $. The new coordinates of the point $ P’ $ are:

$$ ( \cos(2\theta + \frac{\pi}{4}), \sin(2\theta + \frac{\pi}{4}) ) $$

For example, if $ \theta = \frac{\pi}{6} $, then:

$$ \theta’ = 2\times \frac{\pi}{6} + \frac{\pi}{4} = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12} $$

Thus, the new coordinates are:

$$ P’ ( \cos \frac{7\pi}{12}, \sin \frac{7\pi}{12} ) $$