Unit Circle

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:

$$ (0, 1) $$

Find the exact values of sin, cos, and tan at 30 degrees on the unit circle

First, we need to convert $ 30^{\circ} $ to radians:

$$ 30^{\circ} = \frac{\pi}{6} $$

Using the unit circle, the coordinates for $ \frac{\pi}{6} $ are $ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $

From this, we can find:

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

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

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

Determine the value of tan(θ) when sin(θ) = 3/5 and θ is in the first quadrant

Given that $\sin(θ) = \frac{3}{5}$ and $θ$ is in the first quadrant, we can find $\cos(θ)$ using the Pythagorean identity:

$$\sin^2(θ) + \cos^2(θ) = 1$$

Plugging in the given value:

$$\left(\frac{3}{5}\right)^2 + \cos^2(θ) = 1$$

$$\frac{9}{25} + \cos^2(θ) = 1$$

$$\cos^2(θ) = 1 – \frac{9}{25} = \frac{16}{25}$$

Since $θ$ is in the first quadrant, $\cos(θ)$ is positive:

$$\cos(θ) = \frac{4}{5}$$

Now, we can find $\tan(θ)$:

$$\tan(θ) = \frac{\sin(θ)}{\cos(θ)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$$

Therefore, $\tan(θ) = \frac{3}{4}$.

Find the values of tan(θ) at various angles and verify using the unit circle

To find the values of $ \tan(\theta) $ at various angles and verify using the unit circle, we consider the following angles: $ \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} $

1. For $ \theta = \frac{\pi}{4} $:

$$ \tan(\frac{\pi}{4}) = 1 $$

2. For $ \theta = \frac{3\pi}{4} $:

$$ \tan(\frac{3\pi}{4}) = -1 $$

3. For $ \theta = \frac{5\pi}{4} $:

$$ \tan(\frac{5\pi}{4}) = 1 $$

4. For $ \theta = \frac{7\pi}{4} $:

$$ \tan(\frac{7\pi}{4}) = -1 $$

Verification: Using the unit circle, we observe that at these angles, the tangent value is consistent with the coordinates (x, y) where $ \tan(\theta) = \frac{y}{x} $.