Unit Circle

Find the exact values of sin(θ) and cos(θ) for θ = 5π/6 using the unit circle

To find the exact values of $\sin(\theta)$ and $\cos(\theta)$ for $\theta = \frac{5\pi}{6}$, we use the unit circle.

The angle $\frac{5\pi}{6}$ radians is in the second quadrant, where sine is positive and cosine is negative.

The reference angle for $\frac{5\pi}{6}$ is $\pi – \frac{5\pi}{6} = \frac{\pi}{6}$.

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

Therefore, for $\frac{5\pi}{6}$, the coordinates are $\left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)$.

Hence, $\sin(\frac{5\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$.

Convert 120 degrees to radians using the unit circle

To convert degrees to radians, use the formula:

$$\text{radians} = \text{degrees} \times \frac{\pi}{180}$$

Plugging in 120 degrees:

$$120^{\circ} \times \frac{\pi}{180} = \frac{120\pi}{180}$$

Simplify the fraction:

$$\frac{120\pi}{180} = \frac{2\pi}{3}$$

So, 120 degrees is equal to:

$$\frac{2\pi}{3} \text{ radians}$$

Derivation and Memorization Techniques for the Unit Circle

In order to memorize the unit circle, one effective method is to understand how it is derived from fundamental trigonometric principles. Let’s start by deriving key points:

We know the unit circle has a radius of 1. The key angles we need to memorize are $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$.

Calculate the sine and cosine values for $\theta = \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$:

For $\theta = \frac{\pi}{6}$: $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, $\sin(\frac{\pi}{6}) = \frac{1}{2}$

For $\theta = \frac{\pi}{4}$: $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

For $\theta = \frac{\pi}{3}$: $\cos(\frac{\pi}{3}) = \frac{1}{2}$, $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$

By memorizing these values and extending them to other quadrants, we can recall any point on the unit circle:

$$\cos(\theta) = x-coordinate, \sin(\theta) = y-coordinate$$

Thus, the coordinates for $\theta = \frac{\pi}{6}$ are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$, and similar steps apply for other angles.

Find the sine, cosine, and tangent values of the angle 225 degrees using the unit circle

To find the sine, cosine, and tangent values of the angle $225^{\circ}$ using the unit circle, follow these steps:

1. Convert the angle to radians: $$225^{\circ} = 225 \times \frac{\pi}{180} = \frac{5\pi}{4}$$

2. Determine the reference angle: The reference angle for $225^{\circ}$ is $$225^{\circ} – 180^{\circ} = 45^{\circ}$$

3. Use the unit circle to find the coordinates of the reference angle in the third quadrant, where both sine and cosine are negative. The coordinates for $45^{\circ}$ are $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, so for $225^{\circ}$, they will be $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$.

4. From these coordinates, we get:

$$\sin(225^{\circ}) = -\frac{\sqrt{2}}{2}$$

$$\cos(225^{\circ}) = -\frac{\sqrt{2}}{2}$$

$$\tan(225^{\circ}) = \frac{\sin(225^{\circ})}{\cos(225^{\circ})} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1$$

What is the sine of 45 degrees in the unit circle?

First, we need to remember that the unit circle has a radius of 1 and it is centered at the origin (0,0).

To find the sine of 45 degrees, we use the coordinates of the point where the terminal side of the angle intersects the unit circle.

The angle of 45 degrees is in the first quadrant, and for an angle θ in the unit circle, the coordinates of the point are (cosθ, sinθ).

For 45 degrees, which is π/4 radians, both the cosine and sine values are equal to $$ \frac{\sqrt{2}}{2} $$.

Therefore, the sine of 45 degrees is:

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

What is the sine value of an angle of π/4 radians on the unit circle?

To find the sine value of an angle of $\frac{\pi}{4}$ radians on the unit circle, we use the unit circle properties. The angle $\frac{\pi}{4}$ radians is equivalent to 45 degrees.

On the unit circle, the coordinates of the point at $\frac{\pi}{4}$ radians are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$. The sine of the angle is the y-coordinate of this point.

Therefore, $$\sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}$$.

Find the value of tan(θ) given a point on the unit circle

Given a point on the unit circle at coordinates $(x, y)$, find the value of $\tan(\theta)$ where $\theta$ is the angle formed by the radius connecting the point to the origin.

Using the definition of tangent in the unit circle:

$$\tan(\theta) = \frac{y}{x}$$

For example, if the point on the unit circle is $(\frac{1}{2}, \frac{\sqrt{3}}{2})$, then:

$$\tan(\theta) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$$

Determine the Quadrant on a Unit Circle

To determine the quadrant of the angle \( \theta \) on the unit circle, we need to understand the angle’s position in relation to the x-axis and y-axis.

Consider the angle \( \theta = 150^{\circ} \).

Step 1: Convert the angle to radians if needed. \( 150^{\circ} = \frac{5\pi}{6} \) radians.

Step 2: Identify the reference angle and its position. Since \( 150^{\circ} \) is between \( 90^{\circ} \) and \( 180^{\circ} \), it lies in the second quadrant.

Answer: The quadrant of \( 150^{\circ} \) is Quadrant II.

Find the value of cosine for an angle on the unit circle

Let’s find the value of $\cos(\frac{\pi}{4})$ on the unit circle.

The angle $\frac{\pi}{4}$ is equivalent to 45 degrees.

On the unit circle, the coordinates of the point at an angle of $\frac{\pi}{4}$ are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

Thus, the cosine of $\frac{\pi}{4}$ is $$\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$.

Given a point P on the unit circle such that its coordinates are (cos(θ), sin(θ)), find the coordinates of the point Q, which is the reflection of P across the line y = x Then, find the coordinates of the point R, which is the reflection of Q across the

To find the coordinates of the point $Q$, which is the reflection of $P$ across the line $y = x$, we switch the coordinates of $P$. Therefore, the coordinates of $Q$ are $(sin(\theta), cos(\theta))$.

Next, to find the coordinates of the point $R$, which is the reflection of $Q$ across the $x$-axis, we negate the y-coordinate of $Q$. Thus, the coordinates of $R$ are $(sin(\theta), -cos(\theta))$.

Summary:
Coordinates of $Q$: $(sin(\theta), cos(\theta))$
Coordinates of $R$: $(sin(\theta), -cos(\theta))$