Unit Circle

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 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,

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

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 slope of the tangent line to the unit circle at the point where the angle with the positive x-axis is π/3

The unit circle is defined by the equation $ x^2 + y^2 = 1 $.

At the point where the angle with the positive x-axis is $ \pi / 3 $, the coordinates are $ ( \cos (\pi / 3), \sin (\pi / 3 )) $, which simplifies to $ ( \frac{1}{2}, \frac{\sqrt{3}}{2} ) $.

The derivative of the equation $ x^2 + y^2 = 1 $ implicitly gives the slope of the tangent line. Differentiating implicitly with respect to $ x $ gives:

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

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

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

Substituting $ x = \frac{1}{2} $ and $ y = \frac{\sqrt{3}}{2} $:

$$ \frac{dy}{dx} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$

Therefore, the slope of the tangent line is $ -\frac{\sqrt{3}}{3} $.

Find the values of θ in degrees, such that tan(θ) = -1 in the interval [0°, 360°]

To solve for $\theta$ such that $\tan(\theta) = -1$ in the interval $[0°, 360°]$, we first recognize that the tangent function is negative in the second and fourth quadrants.

In the second quadrant, $\tan(180° – \theta) = -1$. So:

$$180° – \theta = 45°$$

Solving for $\theta$:

$$\theta = 180° – 45°$$

$$\theta = 135°$$

In the fourth quadrant, $\tan(360° – \theta) = -1$. So:

$$360° – \theta = 45°$$

Solving for $\theta$:

$$\theta = 360° – 45°$$

$$\theta = 315°$$

Thus, the values of $\theta$ that satisfy $\tan(\theta) = -1$ in the interval $[0°, 360°]$ are:

$$\boxed{135°, 315°}$$

Calculate the sine, cosine and tangent values for angles on the unit circle

Let’s calculate the sine, cosine, and tangent values of the angle $\frac{5\pi}{6}$ on the unit circle.

First, determine the coordinates of the angle $\frac{5\pi}{6}$.

Since $\frac{5\pi}{6} = 180^\circ – 30^\circ$, it is in the second quadrant where sine is positive, and cosine is negative.

Coordinates of $30^\circ$ are $(\cos 30^\circ, \sin 30^\circ) = (\frac{\sqrt{3}}{2}, \frac{1}{2})$.

Therefore, the coordinates of $\frac{5\pi}{6}$ are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

Thus,

$$\sin \left(\frac{5\pi}{6}\right) = \frac{1}{2}$$

$$\cos \left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$

For tangent,

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

In the unit circle, find the coordinates of the point corresponding to an angle of 5π/6 radians Explain your steps and reasoning

To find the coordinates of the point corresponding to an angle of $\frac{5\pi}{6}$ radians on the unit circle, we need to find the cosine and sine of the angle.

First, observe that $\frac{5\pi}{6}$ radians is in the second quadrant.

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

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

Thus, the coordinates are $(-\cos \frac{\pi}{6}, \sin \frac{\pi}{6})$.

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

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