Unit Circle

Find the value of sin(π/3) using the unit circle

$$\text{The angle } \frac{\pi}{3} \text{ is equivalent to } 60^{\circ}.$$

$$\text{On the unit circle, the coordinates for } 60^{\circ} \text{ are } \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$$

$$\sin(\frac{\pi}{3}) \text{ is the y-coordinate, which is } \frac{\sqrt{3}}{2}.$$

Find the sine and cosine of 𝜋/6 radians on the unit circle

To find the sine and cosine of $\frac{\pi}{6}$ radians on the unit circle, we need to recall the standard angle values:

At $\frac{\pi}{6}$ radians:

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

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

Thus, the cosine of $\frac{\pi}{6}$ radians is $\frac{\sqrt{3}}{2}$, and the sine of $\frac{\pi}{6}$ radians is $\frac{1}{2}$.

Find the coordinates and trigonometric values for an angle on the unit circle

Consider an angle $ \theta = \frac{7\pi}{6} $ on the unit circle. We need to find the coordinates of the point on the unit circle corresponding to this angle, as well as the sine and cosine values.

First, identify the reference angle: $$ \theta_{ref} = \pi – \frac{7\pi}{6} = \frac{\pi}{6} $$

Next, find the coordinates for the reference angle $ \frac{\pi}{6} $:

$$ \left( \cos\left(\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right) \right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $$

Since $ \theta = \frac{7\pi}{6} $ is in the third quadrant, both sine and cosine are negative:

$$ \left( \cos\left(\frac{7\pi}{6}\right), \sin\left(\frac{7\pi}{6}\right) \right) = \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$

How to Remember the Unit Circle Fast

$$\text{To remember the unit circle, focus on key angles and their coordinates. Start with } 0^\circ, 30^\circ, 45^\circ, 60^\circ, \text{ and } 90^\circ.$$

$$\text{For example, at } 0^\circ, \text{ the coordinates are } (1, 0).$$

$$\text{At } 30^\circ, \text{ the coordinates are } \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right).$$

$$\text{At } 45^\circ, \text{ the coordinates are } \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right).$$

$$\text{At } 60^\circ, \text{ the coordinates are } \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right).$$

$$\text{At } 90^\circ, \text{ the coordinates are } (0, 1).$$

$$\text{Memorize these points, and use symmetry to fill in the rest of the circle.}$$

Find the value of tan(θ) on the unit circle for θ = 5π/4

To find the value of $ \tan(\theta) $ on the unit circle for $ \theta = \frac{5\pi}{4} $, we first determine the coordinates of the point on the unit circle that corresponds to this angle.

The angle $ \frac{5\pi}{4} $ is in the third quadrant, where both sine and cosine values are negative. The reference angle is $ \pi/4 $.

The coordinates for $ \theta = \frac{5\pi}{4} $ are $ (-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}) $.

Since $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $, we have:

$$ \tan\left(\frac{5\pi}{4}\right) = \frac{\sin\left(\frac{5\pi}{4}\right)}{\cos\left(\frac{5\pi}{4}\right)} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 $$

Find the value of sec(θ) if point P(1/2, √3/2) lies on the unit circle

To find $\sec(\theta)$, we need to know $\cos(\theta)$. Given the coordinates on the unit circle, $\cos(\theta) = x$-coordinate of point $P$.

Here, $x = \frac{1}{2}$. Therefore, $\cos(\theta) = \frac{1}{2}$.

Recall that $\sec(\theta) = \frac{1}{\cos(\theta)}$.

Thus, $\sec(\theta) = \frac{1}{\frac{1}{2}} = 2$.

Therefore, $\sec(\theta) = 2$.

Find the coordinates of the points where the line y = x intersects the unit circle

We start with the unit circle equation:

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

Substituting $y = x$, we get:

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

$$2x^2 = 1$$

$$x^2 = \frac{1}{2}$$

$$x = \pm \frac{\sqrt{2}}{2}$$

Since $y = x$, the coordinates are:

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

and

$$( – \frac{\sqrt{2}}{2}, – \frac{\sqrt{2}}{2})$$

Determine the values of sin(θ) and cos(θ) for θ = 5π/6

Let $θ = \frac{5π}{6}$. This angle is in the second quadrant.

To find $\sin(θ)$ and $\cos(θ)$, we use the reference angle $θ’ = π – \frac{5π}{6} = \frac{π}{6}$.

The sine and cosine of $\frac{π}{6}$ are:

$$\sin\left(\frac{π}{6}\right) = \frac{1}{2}, \cos\left(\frac{π}{6}\right) = \frac{\sqrt{3}}{2}$$

Since the angle is in the second quadrant, $\sin(θ)$ is positive and $\cos(θ)$ is negative.

Thus,

$$\sin\left(\frac{5π}{6}\right) = \frac{1}{2}, \cos\left(\frac{5π}{6}\right) = -\frac{\sqrt{3}}{2}$$

On the unit circle, what is the value of sin(π/6)?

To find the value of $\sin(\frac{\pi}{6})$ on the unit circle, we start by understanding that $\frac{\pi}{6}$ radians is equivalent to 30 degrees. In a unit circle, the coordinates of the angle $\frac{\pi}{6}$ are (\(\frac{\sqrt{3}}{2}, \frac{1}{2}\)).

Therefore, $\sin(\frac{\pi}{6})$ is the y-coordinate, which is:

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

Find the exact values of sin(θ) and cos(θ) for θ = 3π/4 using the unit circle

To find the exact values of $\sin(\theta)$ and $\cos(\theta)$ for $\theta = \frac{3\pi}{4}$, we can use the unit circle.

First, note that $\theta = \frac{3\pi}{4}$ is in the second quadrant. In the unit circle, the angle $\frac{3\pi}{4}$ corresponds to $135^\circ$.

For angles in the second quadrant, the sine value is positive and the cosine value is negative. The reference angle for $\frac{3\pi}{4}$ is $\frac{\pi}{4}$ (or $45^\circ$), where $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

Therefore, $\sin(\frac{3\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{3\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$.

So, the exact values are $\sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$.