Unit Circle

Find the sine and cosine values of the angle $\theta = \frac{\pi}{4}$ on the unit circle

To find the sine and cosine values for $\theta = \frac{\pi}{4}$, we refer to the unit circle.

On the unit circle, the coordinates of the point corresponding to $\theta = \frac{\pi}{4}$ are $\left( \cos \frac{\pi}{4}, \sin \frac{\pi}{4} \right)$.

Since $\frac{\pi}{4}$ is a commonly known angle, we know that:

$$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

$$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

Therefore, the sine and cosine values for $\theta = \frac{\pi}{4}$ are both $\frac{\sqrt{2}}{2}$.

Find the sine, cosine, and tangent values for the angle 45° on the unit circle

To solve for the sine, cosine, and tangent values for the angle $45^{\circ}$ on the unit circle, we use the following properties of the unit circle:

The coordinates for any angle $\theta$ on the unit circle are $(\cos \theta, \sin \theta)$. For $45^{\circ}$, we have:

$$\cos 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

$$\sin 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

To find the tangent value, we use the formula: $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ For $45^{\circ}$, we get: $$\tan 45^{\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$$

On the unit circle, find the coordinates of the point corresponding to the angle $\frac{5\pi}{4}$ radians

To find the coordinates of the point corresponding to the angle $\frac{5\pi}{4}$ radians on the unit circle, we utilize the unit circle’s properties.

First, note that $\frac{5\pi}{4}$ radians is in the third quadrant, where both sine and cosine values are negative.

The reference angle for $\frac{5\pi}{4}$ radians is $\pi – \frac{\pi}{4} = \frac{\pi}{4}$ radians.

For $\frac{\pi}{4}$ radians, the coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

In the third quadrant, both coordinates are negative, so the coordinates for $\frac{5\pi}{4}$ radians are:

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

Find the sine and cosine of 45 degrees and 135 degrees using the unit circle

Using the unit circle, we know that:

$$45^\circ = \frac{\pi}{4}$$

and

$$135^\circ = \frac{3\pi}{4}$$

For $45^\circ$:

$$\sin 45^\circ = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

$$\cos 45^\circ = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

For $135^\circ$:

$$\sin 135^\circ = \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$$

$$\cos 135^\circ = \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$

Find the tangent values for specific angles on the unit circle

To find the tangent values for specific angles on the unit circle, we can use the fact that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.

Let’s find the tangent value for $\frac{\pi}{4}$:

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

We know that:

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

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

Thus,

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

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

Find the value of cot(θ) given that θ is a point on the unit circle where cos(θ) = a and sin(θ) = b

Since $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$, we can use the given values of $\cos(\theta)$ and $\sin(\theta)$.

Given that $\cos(\theta) = a$ and $\sin(\theta) = b$, we plug these into the cotangent formula:

$$\cot(\theta) = \frac{a}{b}$$

Therefore, the value of $\cot(\theta)$ is $\frac{a}{b}$.

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

Given the angle $\theta = \frac{\pi}{4}$, we need to find the value of $\tan(\theta)$.

On the unit circle, the coordinates for $\theta = \frac{\pi}{4}$ are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

The tangent of an angle is given by the ratio of the y-coordinate to the x-coordinate:

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

Thus, the value of $\tan(\frac{\pi}{4})$ is $1$.

Compute tan(4π/3) using the unit circle

To find $\tan \left( \frac{4\pi}{3} \right)$, we use the unit circle.

The angle $\frac{4\pi}{3}$ is in the third quadrant where tangent is positive since both sine and cosine are negative and $\tan\theta = \frac{\sin\theta}{\cos\theta}$

The reference angle for $\frac{4\pi}{3}$ is $\frac{4\pi}{3} – \pi = \frac{\pi}{3}. $

Therefore, $\sin\left(\frac{4\pi}{3}\right) = -\sin\left( \frac{\pi}{3} \right) = -\frac{\sqrt{3}}{2}$ and $\cos\left(\frac{4\pi}{3}\right) = -\cos\left( \frac{\pi}{3} \right) = -\frac{1}{2}.$

Hence,

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

Find the values of cotangent on the unit circle

The cotangent function is defined as the ratio of the cosine of an angle to the sine of the angle: $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$

Given that the angle $\theta$ is located at $\frac{\pi}{4}$ radians on the unit circle, we need to find the value of $\cot \frac{\pi}{4}$.

Using the unit circle values:

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

Therefore,

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

Calculate the points of intersection of the unit circle and the line passing through the origin with a given slope

To find the points of intersection of the unit circle $x^2 + y^2 = 1$ and the line passing through the origin with slope $m$, we start with the equation of the line, $y = mx$.

Substituting $y = mx$ into the unit circle equation:

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

$$x^2 + m^2x^2 = 1$$

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

$$x^2 = \frac{1}{1 + m^2}$$

$$x = \pm \frac{1}{\sqrt{1 + m^2}}$$

Then, using $y = mx$,

$$y = \pm \frac{m}{\sqrt{1 + m^2}}$$

The points of intersection are:

$$(\frac{1}{\sqrt{1 + m^2}}, \frac{m}{\sqrt{1 + m^2}})$$ and $$(\frac{-1}{\sqrt{1 + m^2}}, \frac{-m}{\sqrt{1 + m^2}})$$