Unit Circle

Calculate the sine and cosine values at 45 degrees using the unit circle

To calculate the sine and cosine values at $45^\circ$ using the unit circle, we recognize that a $45^\circ$ angle forms an isosceles right triangle in the unit circle.

The coordinates of the point where the angle intersects the unit circle are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

Therefore, the values are:

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

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

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 $$ \frac{\pi}{3} $$:

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

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

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

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, $\cos(\theta) = \frac{1}{2}$:

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

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, $\tan(θ) = \frac{3}{4}$.