Math

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

Find the value of csc(π/4) using the unit circle

$$ \text{Step 1: Determine the sine of } \frac{\pi}{4} $$

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

$$ \text{Step 2: Use the definition of cosecant: } csc(\theta) = \frac{1}{\sin(\theta)} $$

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

Find the general solution for cotangent of an angle on the unit circle

To find the general solution for $\cot(\theta)$ on the unit circle, first recall the definition of cotangent, which is $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$.

We are looking for the values of $\theta$ where $\cot(\theta) = k$ for some constant $k$.

Since $\cot(\theta)$ is periodic with period $\pi$, the general solution for $\theta$ can be written as:

$$\theta = \arccot(k) + n\pi$$

where $n$ is any integer. This is because the cotangent function repeats every $\pi$ radians. Thus, the full general solution for $\cot(\theta) = k$ can be written as:

$$\theta = \arccot(k) + n\pi \quad \text{for} \; n \in \mathbb{Z}$$

What is the value of sin(π/4) using the unit circle?

To find the value of $ \sin(\frac{\pi}{4}) $ using the unit circle, we need to identify the coordinates of the point on the unit circle corresponding to $ \frac{\pi}{4} $ radians.

The angle $ \frac{\pi}{4} $ radians is equivalent to 45 degrees. On the unit circle, the coordinates of the point that corresponds to 45 degrees are $ (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $.

Since $ \sin(\theta) $ is equal to the y-coordinate of the point on the unit circle, we have

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

Find the sine of an angle in radians on the unit circle

To find the sine of an angle on the unit circle, we need to find the y-coordinate of the point where the terminal side of the angle intersects the unit circle.

Let’s consider the angle \( \frac{5\pi}{6} \).

First, we need to identify the reference angle, which is the acute angle formed with the x-axis. For \( \frac{5\pi}{6} \), the reference angle is \( \pi – \frac{5\pi}{6} = \frac{\pi}{6} \).

On the unit circle, the coordinates of the point corresponding to \( \frac{\pi}{6} \) are \( ( \frac{\sqrt{3}}{2}, \frac{1}{2} ) \).

Since \( \frac{5\pi}{6} \) is in the second quadrant, the sine value (y-coordinate) remains positive.

Therefore, the sine of \( \frac{5\pi}{6} \) is:

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

Find the Coordinates of a Point on the Unit Circle

Given a unit circle, find the coordinates of the point where the terminal side of an angle $\theta = \frac{5\pi}{4}$ intersects the circle.

Solution:

To find the coordinates of the point where the terminal side of an angle $\theta = \frac{5\pi}{4}$ intersects the unit circle, we use the following formulas:
$x = \cos(\theta)$
$y = \sin(\theta)$

For $\theta = \frac{5\pi}{4}$:
$x = \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$y = \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

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