Homework

Convert 5π/6 radians to degrees and find the sine and cosine values on the unit circle

First, we convert radians to degrees:

$$\text{Degrees} = \frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = 150^{\circ}$$

Next, we identify the sine and cosine values for $150^{\circ}$ on the unit circle:

$$\sin(150^{\circ}) = \sin(180^{\circ} – 30^{\circ}) = \sin(30^{\circ}) = \frac{1}{2}$$

$$\cos(150^{\circ}) = \cos(180^{\circ} – 30^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2}$$

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 $-\frac{\sqrt{3}}{3}$.