Math

Easy Ways to Remember the Unit Circle

One of the easiest ways to remember the unit circle is by understanding the key angles and their coordinates. The unit circle has a radius of 1, and key angles are 0°, 30°, 45°, 60°, 90°, etc. These angles correspond to specific coordinates on the circle:

$$0°: (1, 0)$$

$$30°: \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$

$$45°: \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$$

$$60°: \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$$

$$90°: (0, 1)$$

Memorize these coordinates and their symmetries to quickly recall the unit circle.

Solve for the exact values of sine and cosine for the angle 7π/6 on the unit circle

First, note that $\frac{7\pi}{6}$ radians is in the third quadrant where both sine and cosine are negative. The reference angle for $\frac{7\pi}{6}$ is $\pi – \frac{7\pi}{6} = \frac{\pi}{6}$.

The sine and cosine values for $\frac{\pi}{6}$ are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, respectively. Since $\frac{7\pi}{6}$ is in the third quadrant, these values become negative.

Therefore, the exact values are:

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

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

Find the cosine of the angle formed by the complex number in the unit circle

Given a complex number $z = \cos(\theta) + i\sin(\theta)$ on the unit circle, find the cosine of the angle $\theta$.

We know that for a complex number on the unit circle, $z = e^{i\theta} = \cos(\theta) + i\sin(\theta)$.

The real part of $z$ is $\cos(\theta)$.

Therefore, the cosine of the angle $\theta$ is simply the real part of $z$ which is $\cos(\theta)$.

$$ \cos(\theta) = \Re(z) $$

Find the sine and cosine of 135 degrees using the unit circle

$135^\circ$ is in the second quadrant. The reference angle is $180^\circ – 135^\circ = 45^\circ$.

The sine and cosine values for $45^\circ$ are $ \frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$ respectively.

Since $135^\circ$ is in the second quadrant, the cosine value is negative and the sine value is positive.

Thus, $\cos 135^\circ = -\frac{1}{\sqrt{2}}$ and $\sin 135^\circ = \frac{1}{\sqrt{2}}$.

Find the coordinates of the point on the unit circle at an angle of \( \frac{\pi}{3} \) radians

To find the coordinates of the point on the unit circle at an angle of $ \frac{\pi}{3} $ radians, we use the formula for the coordinates on the unit circle: $ ( \cos \theta, \sin \theta ) $.

Here, $ \theta = \frac{\pi}{3} $.

So, we need to find $ \cos \frac{\pi}{3} $ and $ \sin \frac{\pi}{3} $.

From trigonometric values, we know that:

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

$$ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} $$

Therefore, the coordinates of the point are $$ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $$.

Determine the coordinates on the unit circle for the angle whose cosine is -2/3

To find the coordinates on the unit circle for the angle whose cosine is $-\frac{2}{3}$, we use the Pythagorean identity:

$$\cos^2\theta + \sin^2\theta = 1$$

Given that $\cos\theta = -\frac{2}{3}$, we substitute this into the identity:

$$\left(-\frac{2}{3}\right)^2 + \sin^2\theta = 1$$

This gives:

$$\frac{4}{9} + \sin^2\theta = 1$$

Subtract $\frac{4}{9}$ from both sides:

$$\sin^2\theta = 1 – \frac{4}{9}$$

$$\sin^2\theta = \frac{9}{9} – \frac{4}{9}$$

$$\sin^2\theta = \frac{5}{9}$$

Taking the square root of both sides, we get:

$$\sin\theta = \pm\frac{\sqrt{5}}{3}$$

So, the coordinates on the unit circle are:

$$\left(-\frac{2}{3},\frac{\sqrt{5}}{3}\right) \text{and} \left(-\frac{2}{3},-\frac{\sqrt{5}}{3}\right)$$

Find the slope of the tangent line to the unit circle at the point (1/2, sqrt(3)/2)

To find the slope of the tangent line to the unit circle at a given point, we need to differentiate the equation of the unit circle. The unit circle is defined by:

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

We implicitly differentiate with respect to $x$:

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

Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = -\frac{x}{y}.$$

Next, we substitute the point $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$ into the equation:

$$\frac{dy}{dx} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$$

Therefore, the slope of the tangent line at the point $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$ is $-\frac{\sqrt{3}}{3}$.

Find the cosine of an angle of \(\frac{\pi}{3}\) radians on the unit circle

To find the cosine of an angle of $\frac{\pi}{3}$ radians on the unit circle, we need to look at the coordinates of the point where the terminal side of the angle intersects the unit circle.

On the unit circle, the coordinates are given by $(\cos \theta, \sin \theta)$ where $\theta$ is the angle in radians.

For $\theta = \frac{\pi}{3}$, the coordinates are $(\frac{1}{2}, \frac{\sqrt{3}}{2})$.

Thus, the cosine of $\frac{\pi}{3}$ is $\frac{1}{2}$.

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