Unit Circle

Determine the sine value at an angle of π/4 on the unit circle

To determine the sine value at an angle of $ \frac{\pi}{4} $ on the unit circle, recall that at this angle, the coordinates are:

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

The sine value corresponds to the y-coordinate:

$$ \sin( \frac{\pi}{4} ) = \frac{\sqrt{2}}{2} $$

Determine tan(θ) from the unit circle at point P(x,y)

To determine $ \tan(\theta) $ from the unit circle at point $ P(x,y) $, recall that

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

On the unit circle, you have $ P(x,y) = (\cos(\theta), \sin(\theta)) $, so

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

Ensure that $ x \neq 0 $ to avoid division by zero.

Create a colorful circle pattern using points on the unit circle with $cos(\theta)$ and $sin(\theta)$

To create a colorful circle pattern, you can use points on the unit circle defined by $\cos(\theta)$ and $\sin(\theta)$ where $0 \leq \theta \leq 2\pi$. Each point coordinates can be calculated as:

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

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

For instance, if you plot points for $\theta$ in multiples of $\frac{\pi}{6}$, you will get 12 equally spaced points around a circle.

Determine the coordinates on the unit circle for the angle -2/3π

To determine the coordinates on the unit circle for the angle $-\frac{2}{3}π$, we first convert this angle to its corresponding positive angle by adding $2π$:

$$ -\frac{2}{3}π + 2π = \frac{4π}{3} $$

Now, we find the coordinates corresponding to the angle $\frac{4π}{3}$ on the unit circle. This angle is in the third quadrant, where both sine and cosine are negative:

$$ \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right) $$

Find the value of arcsin(1/2)

To find the value of $ \arcsin(\frac{1}{2}) $, we need to determine the angle $ \theta $ whose sine is $ \frac{1}{2} $.

From the unit circle, we know:

$$ \sin(\theta) = \frac{1}{2} $$

The angle $ \theta $ that satisfies this in the range $ [-\frac{\pi}{2}, \frac{\pi}{2}] $ is:

$$ \theta = \frac{\pi}{6} $$

Thus, $ \arcsin(\frac{1}{2}) = \frac{\pi}{6} $.

Find the coordinates on the unit circle corresponding to an angle theta

To find the coordinates on the unit circle corresponding to an angle $ \theta $ , we use the parametric equations of the unit circle:

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

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

Thus, the coordinates are given by:

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

How to find the reference angle for an angle not located on the unit circle

To find the reference angle for an angle not located on the unit circle, we first need to understand that the reference angle is the smallest angle between the given angle and the x-axis. Let\

What is the value of cos(-π/3) using the unit circle?

To find the value of $\cos(-\frac{\pi}{3})$ using the unit circle, first recognize that the cosine function is an even function. This means that:

$$\cos(-x) = \cos(x)$$

Therefore:

$$\cos(-\frac{\pi}{3}) = \cos(\frac{\pi}{3})$$

From the unit circle, we know that:

$$\cos(\frac{\pi}{3}) = \frac{1}{2}$$

Thus:

$$\cos(-\frac{\pi}{3}) = \frac{1}{2}$$

Find the coordinates of the point on the unit circle at which the angle is 7π/6

To find the coordinates of the point on the unit circle at which the angle is $ \frac{7\pi}{6} $, we use the following:

The unit circle has the equation:

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

The coordinates of a point on the unit circle are given by:

$$ (\cos(\theta), \sin(\theta)) $$

For $ \theta = \frac{7\pi}{6} $:

$$ \cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2} $$

$$ \sin(\frac{7\pi}{6}) = -\frac{1}{2} $$

Therefore, the coordinates are:

$$ \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$

Find the exact coordinates of the point(s) on the unit circle where the tangent line is vertical

The equation of the unit circle is given by:

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

We find the tangent line to be vertical when the derivative is undefined. Thus, we need to find the points where $ \x0crac{dy}{dx} $ is undefined.

Implicitly differentiate the unit circle equation with respect to $ x $:

$$ 2x + 2y \x0crac{dy}{dx} = 0 $$

Simplify and solve for $ \x0crac{dy}{dx} $:

$$ \x0crac{dy}{dx} = -\x0crac{x}{y} $$

The derivative is undefined when $ y = 0 $. Thus, we solve for $ x $:

When $ y = 0 $, substituting back into the original equation:

$$ x^2 = 1 $$

So, $ x = 1 $ or $ x = -1 $.

Therefore, the points are:

$(1,0)$ and $(-1,0)$.