Math

Given a point on the unit circle, determine the coordinates and verify the trigonometric identities

Let’s consider a point $P(\cos\theta, \sin\theta)$ on the unit circle where $\theta = \frac{5\pi}{6}$. To find the coordinates and verify trigonometric identities:

First, we calculate the coordinates:

$$P = (\cos \frac{5\pi}{6}, \sin \frac{5\pi}{6})$$

Using the unit circle, we know:

$$\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$$

$$\sin \frac{5\pi}{6} = \frac{1}{2}$$

Thus, the coordinates are:

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

Next, we verify the Pythagorean identity:

$$\cos^2 \theta + \sin^2 \theta = 1$$

Substituting in the values, we get:

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

Which confirms that the point lies on the unit circle.

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

To find the cosine of the angle \( \frac{\pi}{3} \) on the unit circle, we need to locate this angle on the circle.

The angle \( \frac{\pi}{3} \) corresponds to 60 degrees.

On the unit circle, the coordinates of the point at angle \( \frac{\pi}{3} \) are \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \).

The cosine of an angle is the x-coordinate of the corresponding point on the unit circle.

Therefore, \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \).

Find all angles θ between 0 and 2π such that cos(θ) = -1/2

To find the angles $\theta$ such that $\cos(\theta) = -\frac{1}{2}$, we start by identifying the quadrants where $\cos(\theta)$ is negative. Cosine is negative in the second and third quadrants.

First, we find the reference angle:

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

Now, we find the angles in the second and third quadrants:

Second quadrant: $$\pi – \frac{\pi}{3} = \frac{2\pi}{3}$$

Third quadrant: $$\pi + \frac{\pi}{3} = \frac{4\pi}{3}$$

Thus, the angles are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.

Find the coordinates of a point on the unit circle at an angle of 45 degrees from the positive x-axis

To find the coordinates of a point on the unit circle, we use the trigonometric functions sine and cosine.

The angle given is $45^{\circ}$.

Using the unit circle properties:

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

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

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

Find the value of x such that cos(x) = -1/2 and sin(x) is negative on the unit circle

To solve for $x$ such that $\cos(x) = -\frac{1}{2}$ and $\sin(x)$ is negative on the unit circle, follow these steps:

1. Identify the angles where $\cos(x) = -\frac{1}{2}$. This occurs at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ in radians.

2. Determine which of these angles has a negative sine value. The sine function is negative in the third and fourth quadrants.

3. Since $\frac{2\pi}{3}$ is in the second quadrant and $\frac{4\pi}{3}$ is in the third quadrant, we choose $x = \frac{4\pi}{3}$.

Therefore, the value of $x$ is:

$$ x = \frac{4\pi}{3} $$

Find the exact values of sine and cosine for the angle 5π/4 using the unit circle

To find the exact values of sine and cosine for the angle $\frac{5\pi}{4}$, we start by determining in which quadrant the angle lies.

The angle $\frac{5\pi}{4}$ is in the third quadrant because $\frac{5\pi}{4} > \pi$ and $\frac{5\pi}{4} < \frac{3\pi}{2}$.

In the third quadrant, both sine and cosine are negative.

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

We know that $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

Hence, for the third quadrant:

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

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

Given the point P on the unit circle at an angle of 210 degrees, find cos(210°) and sin(210°)

To find $\cos(210^{\circ})$ and $\sin(210^{\circ})$, we start by converting the angle to radians:

$$210^{\circ} = 210 \cdot \frac{\pi}{180} = \frac{7\pi}{6}$$

The reference angle for $\frac{7\pi}{6}$ is $30^{\circ}$ or $\frac{\pi}{6}$.

The coordinates of a point on the unit circle at an angle $\theta$ in the third quadrant are $(-\cos(\theta), -\sin(\theta))$.

Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$, we get:

$$\cos(210^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2}$$

$$\sin(210^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}$$

Therefore, the coordinates of the point are:

$$P(\cos(210^{\circ}), \sin(210^{\circ})) = \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$

Find the coordinates of the points where the unit circle intersects the x-axis

$$\text{The unit circle has the equation } x^2 + y^2 = 1.$$

$$\text{To find the intersection with the x-axis, we set } y = 0.$$

$$x^2 + 0^2 = 1$$

$$x^2 = 1$$

$$x = \pm 1.$$

$$\text{Thus, the coordinates are } (1, 0) \text{ and } (-1, 0).$$

Given a point P on the unit circle at an angle θ, find the coordinates of P, the length of the line segment from P to the origin, and the area of the sector formed by the angle θ in the unit circle

Given a point $P$ on the unit circle at an angle $\theta$, we can determine the coordinates of $P$ as follows:

$$ P(\cos(\theta), \sin(\theta)) $$

The length of the line segment from $P$ to the origin is simply the radius of the unit circle, which is 1.

To find the area of the sector formed by the angle $\theta$, we use the formula for the area of a sector, $$ A = \frac{1}{2} r^2 \theta $$ Since the radius $r$ is 1,

$$ A = \frac{1}{2} \theta $$

Therefore, the coordinates of $P$ are $(\cos(\theta), \sin(\theta))$, the length of the line segment from $P$ to the origin is 1, and the area of the sector is $\frac{1}{2} \theta$.

Find the angle that corresponds to a given point on the unit circle

Let’s consider the point (\frac{\sqrt{3}}{2}, \, \frac{1}{2}) on the unit circle. This point lies in the first quadrant and has coordinates (cos(\theta), sin(\theta)). We need to find the angle \theta that corresponds to this point.

Using the coordinates, we know that

$$ \cos(\theta) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(\theta) = \frac{1}{2} $$

The angle \theta that satisfies both these conditions is

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

Therefore, the angle corresponding to the point (\frac{\sqrt{3}}{2}, \frac{1}{2}) is \frac{\pi}{6} radians.