Unit Circle

Find the coordinates of cos(π/3) on the unit circle

To find the coordinates of $ \cos(\frac{\pi}{3}) $ on the unit circle, we need to identify the coordinates associated with this angle.

On the unit circle, the angle $ \frac{\pi}{3} $ corresponds to the 60° position.

At this position, the coordinates are:

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

Determine the general solution for sin(x) = 1/2 within [0, 2π]

To determine the general solution for $ \sin(x) = \frac{1}{2} $ within the interval $ [0, 2\pi] $, we need to find all the angles $ x $ where the sine function yields $ \frac{1}{2} $ on the unit circle.

The sine function equals $ \frac{1}{2} $ at angles $ \frac{\pi}{6} $ and $ \frac{5\pi}{6} $ within the given interval.

Thus, the general solutions are:

$$ x = \frac{\pi}{6} + 2n\pi $$

and

$$ x = \frac{5\pi}{6} + 2n\pi $$

where $ n $ is any integer.

What are the coordinates of 3π/4 on the unit circle?

The coordinates of $ \frac{3\pi}{4} $ on the unit circle can be found using the unit circle definitions. The angle $ \frac{3\pi}{4} $ corresponds to $ 135^{\circ} $. At this angle, the coordinates are:

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

Find the sine and cosine of 7π/6 on the unit circle

To find the sine and cosine of $ \frac{7\pi}{6} $ on the unit circle, we first determine the reference angle. The reference angle for $ \frac{7\pi}{6} $ is $ \frac{\pi}{6} $.

The sine and cosine of $ \frac{7\pi}{6} $ correspond to the sine and cosine of $ \frac{\pi}{6} $ but with signs corresponding to the third quadrant.

From the unit circle, we know:

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

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

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

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

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

Find the value of tan at π/4 on the unit circle

To find the value of $ \tan(\frac{\pi}{4}) $ on the unit circle, we use the definition of tangent, which is the ratio of sine to cosine:

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

At $ \theta = \frac{\pi}{4} $, both $ \sin(\frac{\pi}{4}) $ and $ \cos(\frac{\pi}{4}) $ are equal to $ \frac{\sqrt{2}}{2} $:

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

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.

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 $(-1,0)$.

Find the value of sin(θ), cos(θ), and tan(θ) for θ = π/3 on the unit circle

When $θ = \fracπ3$, we can find the values of $\sin(θ)$, $\cos(θ)$, and $\tan(θ)$ from the unit circle:

$$\sin(\fracπ3) = \frac{\sqrt3}2$$

$$\cos(\fracπ3) = \frac12$$

$$\tan(\fracπ3) = \frac{\sin(\fracπ3)}{\cos(\fracπ3)} = \sqrt3$$

Identify the coordinates of the point on the unit circle at an angle of π/4

On the unit circle, the coordinates of the point at an angle of $ \frac{\pi}{4} $ are:

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

Determine the exact values of tan(θ) for θ = 5π/6, θ = 3π/4, and θ = 7π/4 from the unit circle

To determine the exact values of $ \tan(\theta) $ for the given angles using the unit circle, we need to recall the tangent function and its relation to sine and cosine:

\n

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

\n

1. For $ \theta = \frac{5\pi}{6} $:

\n

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

\n

Therefore:

\n

$$ \tan(\frac{5\pi}{6}) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$

\n

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

\n

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

\n

Therefore:

\n

$$ \tan(\frac{3\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

\n

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

\n

$$ \sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2}, \quad \cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2} $$

\n

Therefore:

\n

$$ \tan(\frac{7\pi}{4}) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$