Math

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

What is the value of sin(π/6) and cos(π/6) on the unit circle?

To find the values of $\sin(\frac{\pi}{6})$ and $\cos(\frac{\pi}{6})$ on the unit circle, we need to understand the coordinates of the point on the unit circle corresponding to the angle $\frac{\pi}{6}$. The unit circle has a radius of 1, and the coordinates at an angle $\theta$ are $(\cos(\theta), \sin(\theta))$.

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

The coordinates on the unit circle are $\left(\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})\right)$.

From the unit circle chart:

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

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

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

Find the exact value of tan(θ) given the point on the unit circle

Given a point on the unit circle at $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$, find the exact value of $\tan(\theta)$.

First, identify the coordinates $x$ and $y$ from the point, which are $x = -\frac{1}{2}$ and $y = -\frac{\sqrt{3}}{2}$ respectively. Recall that $\tan(\theta) = \frac{y}{x}$.

Plug in the values of $x$ and $y$:

$$\tan(\theta) = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$

Therefore, the exact value of $\tan(\theta)$ is $\sqrt{3}$.

Find the coordinates on the unit circle for the angle corresponding to -2/3π radians

To find the coordinates on the unit circle for the angle corresponding to $-\frac{2}{3}\pi$ radians, we first determine the reference angle. The reference angle is $\frac{2}{3}\pi$ radians.

The coordinates for $\frac{2}{3}\pi$ radians are $(\cos(\frac{2}{3}\pi), \sin(\frac{2}{3}\pi))$.

Calculating these values, we get:

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

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

Since the angle is negative, the coordinates will be in the third quadrant, so both values will be negative:

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

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

Therefore, the coordinates are: $$(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$$

Find the value of secant on the unit circle for an angle of 60 degrees

First, recall that $\sec \theta = \frac{1}{\cos \theta}$.

For $\theta = 60^\circ$, we know from the unit circle that $\cos 60^\circ = \frac{1}{2}$.

Therefore,

$$\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2.$$