Homework

Find the angle corresponding to the point (1/2, -√3/2) on the unit circle

To find the angle that corresponds to the point $ \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) $ on the unit circle, we look at the coordinates.

The x-coordinate is $ \frac{1}{2} $ and the y-coordinate is $ -\frac{\sqrt{3}}{2} $. These values correspond to an angle in the fourth quadrant.

The reference angle with these coordinates is $ \frac{\pi}{3} $ because:

$$ \cos \theta = \frac{1}{2} \text{ and } \sin \theta = -\frac{\sqrt{3}}{2} $$

Since the angle is in the fourth quadrant, the actual angle is:

$$ \theta = 2\pi – \frac{\pi}{3} = \frac{5\pi}{3} $$

Find the values of θ for which tan(θ) = 1 in the unit circle

To find the values of $ \theta $ for which $ \tan(\theta) = 1 $ on the unit circle, we need to identify the angles where the tangent function is equal to 1.

The tangent function is defined as the ratio of the sine and cosine functions:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$

For $ \tan(\theta) = 1 $, we have:

$$ \frac{\sin(\theta)}{\cos(\theta)} = 1 $$

This implies:

$$ \sin(\theta) = \cos(\theta) $$

On the unit circle, this equality occurs at:

$$ \theta = \frac{\pi}{4} + k\pi $$

where $ k $ is any integer. Therefore, the solutions are:

$$ \theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \ldots $$

Find the exact values of the trigonometric functions for the angle 5π/3 in the unit circle

To find the exact values of the trigonometric functions for the angle $$ \frac{5\pi}{3} $$ in the unit circle, we first note that:

$$ \frac{5\pi}{3} = 2\pi – \frac{\pi}{3} $$

This means the angle is located in the fourth quadrant.

We can use the reference angle of $$ \frac{\pi}{3} $$:

$$ \cos \left( \frac{5\pi}{3} \right) = \cos \left( 2\pi – \frac{\pi}{3} \right) = \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} $$

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

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

Identify the location of -pi/2 on a unit circle

On the unit circle, angles are measured starting from the positive x-axis and moving counterclockwise.

To find the location of $-\pi/2$:

– Starting from the positive x-axis, move clockwise because the angle is negative.

– $\pi/2$ is 90 degrees, so moving clockwise 90 degrees lands you on the negative y-axis.

Therefore, the coordinates of $-\pi/2$ on the unit circle are:

$ (0, -1) $

Determine the value of theta for which the point (cos(θ), sin(θ)) on the unit circle forms a right-angled triangle with the origin and the point (1, 0)

Given the points $ (\cos(\theta), \sin(\theta))$, the origin $(0, 0)$, and $(1, 0)$, we need to find $ \theta $ such that they form a right-angled triangle.

The distance between $(\cos(\theta), \sin(\theta))$ and $(1, 0)$ is:

$$ d = \sqrt{(\cos(\theta) – 1)^2 + \sin^2(\theta)} $$

Since $(\cos(\theta), \sin(\theta))$ lies on the unit circle, we use the Pythagorean identity:

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

Thus the distance simplifies to:

$$ d = \sqrt{1 – 2\cos(\theta) + 1} = \sqrt{2 – 2\cos(\theta)} = \sqrt{2(1 – \cos(\theta))} $$

For the triangle to be right-angled, $\cos(\theta) = \frac{1}{2}$:

$$ \cos(\theta) = \frac{1}{2} \implies \theta = \frac{\pi}{3} \text{ or } \theta = -\frac{\pi}{3} $$

Calculate the area of a sector in a unit circle with a given central angle θ

To calculate the area of a sector in a unit circle with a given central angle $ \theta $ (in radians), use the following formula:

$$ A = \frac{1}{2} \cdot r^2 \cdot \theta $$

Since the radius $ r $ of a unit circle is 1, the formula simplifies to:

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

So, the area of the sector is:

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

Find the values of tan(θ), sin(θ), and cos(θ) for θ = 45 degrees

To find the values of $\tan(\theta)$, $\sin(\theta)$, and $\cos(\theta)$ for $\theta = 45^\circ$:

First, we note that $\theta = 45^\circ$ is in the first quadrant of the unit circle.

The coordinates of the point on the unit circle at $\theta = 45^\circ$ are:

$$\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$$

Therefore:

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

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

Using the definition of tangent:

$$\tan(45^\circ) = \frac{\sin(45^\circ)}{\cos(45^\circ)} = 1$$

Prove the identity involving cos and sin on the unit circle

To prove the identity involving $ \cos(\theta) $ and $ \sin(\theta) $ on the unit circle, we start with the Pythagorean identity:

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

Consider the parameterization of the unit circle with $ \theta $ as the angle from the positive x-axis:

$$ x = \cos(\theta) $$

$$ y = \sin(\theta) $$

Then, the coordinates $ (x, y) $ must satisfy:

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

Substituting $ x = \cos(\theta) $ and $ y = \sin(\theta) $, we get:

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

This verifies the identity.

Find the coordinates of the point where the line y = 1 intersects the unit circle

To find the coordinates where the line $ y = 1 $ intersects the unit circle, we start by recalling the equation of the unit circle:

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

Substituting $ y = 1 $ into the unit circle equation, we get:

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

Simplifying,

$$ x^2 + 1 = 1 $$

$$ x^2 = 0 $$

$$ x = 0 $$

Therefore, the point of intersection is:

$$ (0, 1) $$