Math

Determine the angle measures of a point on the unit circle

Given a point $P$ on the unit circle with coordinates $P = (\frac{3}{5}, -\frac{4}{5})$, determine all possible angle measures $\theta$ in degrees.

First, we calculate the reference angle $\alpha$ by using the trigonometric functions. Notice that the coordinates of $P$ give us the cosine and sine of $\theta$:

$$\cos(\theta) = \frac{3}{5}, \sin(\theta) = -\frac{4}{5}$$

Using the inverse cosine function, we find the reference angle:

$$\alpha = \cos^{-1}(\frac{3}{5}) \approx 53.13^\circ$$

Since the sine is negative and the cosine is positive, $\theta$ is in the fourth quadrant. Therefore,

$$ \theta = 360^\circ – \alpha = 360^\circ – 53.13^\circ \approx 306.87^\circ$$

Thus, the possible angle measures are:

$$ \theta \approx 306.87^\circ $$

Finding Specific Trigonometric Values on the Unit Circle

Given the angle θ = 225°, find the sine, cosine, and tangent values using the unit circle.

First, convert the angle 225° to radians: $$225° = \frac{5\pi}{4}$$ radians.

On the unit circle, the angle $$\frac{5\pi}{4}$$ is located in the third quadrant, where sine and cosine are both negative.

The coordinates for $$\frac{5\pi}{4}$$ are $$(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$$.

Thus, $$\sin(225°) = -\frac{\sqrt{2}}{2}$$, $$\cos(225°) = -\frac{\sqrt{2}}{2}$$

Finally, the tangent value: $$\tan(225°) = \frac{\sin(225°)}{\cos(225°)} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1$$.

Therefore, $$\sin(225°) = -\frac{\sqrt{2}}{2}$$, $$\cos(225°) = -\frac{\sqrt{2}}{2}$$, $$\tan(225°) = 1$$.

Find the tangent of the angle

Given an angle \( \theta = \frac{\pi}{4} \), find \( \tan(\theta) \).

Since the angle \( \theta \) is within the first quadrant and \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we have:

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

Therefore:

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

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

Find the cotangent of the angle $\theta = \frac{\pi}{4}$ on the unit circle

To find the cotangent of $\theta = \frac{\pi}{4}$ on the unit circle:

The cotangent function is given by:

$$\cot \theta = \frac{1}{\tan \theta}$$

Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we first find the values of $\sin \theta$ and $\cos \theta$. For $\theta = \frac{\pi}{4}$:

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

Then:

$$\tan \frac{\pi}{4} = \frac{\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}} = 1$$

Thus, the cotangent is:

$$\cot \frac{\pi}{4} = \frac{1}{\tan \frac{\pi}{4}} = 1$$

So, the answer is:

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

Given a point on the unit circle at an angle of 5π/4 radians, find the coordinates of this point Then, if the unit circle is flipped about the y-axis, determine the new coordinates of the original point after the flip

First, find the coordinates of the point on the unit circle at $\frac{5\pi}{4}$ radians. This point can be represented as:

$$ (\cos(\frac{5\pi}{4}), \sin(\frac{5\pi}{4})) $$

We know that:

$$ \cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} $$

$$ \sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} $$

Thus, the coordinates at $\frac{5\pi}{4}$ radians are:

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

Next, when the unit circle is flipped about the y-axis, the x-coordinate of the point changes sign, but the y-coordinate remains the same. Therefore, the new coordinates after the flip are:

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

Find the values of cotangent for a given angle on the unit circle and verify their consistency

To find the values of $\cot(\theta)$ for $\theta = \frac{3\pi}{4}$ on the unit circle, we start by identifying the coordinates of this angle on the unit circle.

The coordinates for $\theta = \frac{3\pi}{4}$ are $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. The cotangent function is defined as the cosine divided by the sine of the angle: $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$.

Therefore,

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

The value of $\cot\left(\frac{3\pi}{4}\right)$ is -1.

What is the cosine of the angle at 2π/3 on the unit circle?

First, recognize that the angle $ \frac{2\pi}{3} $ is in the second quadrant.

In the unit circle, the cosine function is negative in the second quadrant.

The reference angle for $ \frac{2\pi}{3} $ is $ \pi – \frac{2\pi}{3} = \frac{\pi}{3} $.

We know that $ \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} $.

Therefore, the cosine of $ \frac{2\pi}{3} $ is:

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

Determine the Tangent Slope at a Given Point on the Unit Circle

Let the given point on the unit circle be $(a, b)$, where $a^2 + b^2 = 1$. We need to determine the slope of the tangent line at this point.

The equation of the unit circle is given by:

$$x^2 + y^2 = 1$$

To find the slope of the tangent line at $(a, b)$, we first implicitly differentiate both sides of the equation with respect to $x$:

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

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

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

Substituting the point $(a, b)$ into the derivative:

$$\frac{dy}{dx}\bigg|_{(a,b)} = -\frac{a}{b}$$

Therefore, the slope of the tangent line at the point $(a, b)$ is $-\frac{a}{b}$.

Find the value of tan(4π/3) using the unit circle

$$To \ find \ the \ value \ of \ tan(\frac{4\pi}{3}) \ using \ the \ unit \ circle, \ we \ first \ need \ to \ determine \ the \ reference \ angle. \ $$

$$The \ angle \ \frac{4\pi}{3} \ is \ in \ the \ third \ quadrant, \ and \ its \ reference \ angle \ is \ \frac{4\pi}{3} – \pi = \frac{\pi}{3}. \ $$

$$Using \ the \ unit \ circle, \ the \ coordinates \ for \ \frac{\pi}{3} \ are \ (\frac{1}{2}, \ \frac{\sqrt{3}}{2}). \ $$

$$Since \ the \ angle \ \frac{4\pi}{3} \ is \ in \ the \ third \ quadrant, \ both \ x \ and \ y \ coordinates \ are \ negative: \ (-\frac{1}{2}, \ -\frac{\sqrt{3}}{2}). \ $$

$$Finally, \ the \ tangent \ is \ the \ ratio \ of \ y \ to \ x: \ tan(\frac{4\pi}{3}) = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}. \ $$

$$Therefore, \ tan(\frac{4\pi}{3}) = \sqrt{3}. \ $$

Finding the Location of -π/2 on a Unit Circle

To find the location of $-\pi/2$ on the unit circle, we start by understanding that angles are measured from the positive x-axis, and negative angles are measured clockwise.

For $-\pi/2$ radians, start from the positive x-axis and measure clockwise by $\pi/2$ radians (or 90 degrees). This brings us to the negative y-axis.

The coordinates of this point on the unit circle are $$(0, -1)$$.

So, $-\pi/2$ radians corresponds to the point (0, -1) on the unit circle.