Homework

Find the cosine of θ if sin(θ) = 1/2 and θ is in the first quadrant

Given $ \sin(\theta) = \frac{1}{2} $ and $ \theta $ is in the first quadrant.

We know that $ \sin^2(\theta) + \cos^2(\theta) = 1 $.

So,

$$ \left( \frac{1}{2} \right)^2 + \cos^2(\theta) = 1 $$

$$ \frac{1}{4} + \cos^2(\theta) = 1 $$

$$ \cos^2(\theta) = 1 – \frac{1}{4} $$

$$ \cos^2(\theta) = \frac{3}{4} $$

$$ \cos(\theta) = \pm \sqrt{\frac{3}{4}} $$

Since $ \theta $ is in the first quadrant, $ \cos(\theta) $ is positive:

$$ \cos(\theta) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $$

Calculate the exact values of the trigonometric functions for an angle of 7π/6 radians on the unit circle

To find the trigonometric functions for the angle $ \frac{7\pi}{6} $, locate the angle on the unit circle.

First, convert $ \frac{7\pi}{6} $ to degrees: $$ \frac{7\pi}{6} \times \frac{180^\circ}{\pi} = 210^\circ $$

Next, find the reference angle: $$ 210^\circ – 180^\circ = 30^\circ $$

Using the reference angle and the unit circle values, we have:

$$ \sin\left(\frac{7\pi}{6}\right) = -\sin\left(30^\circ\right) = -\frac{1}{2} $$

$$ \cos\left(\frac{7\pi}{6}\right) = -\cos\left(30^\circ\right) = -\frac{\sqrt{3}}{2} $$

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

Find the tangent value of π/4 in the unit circle

To find the tangent value of $ \frac{\pi}{4} $ in the unit circle, use the definition of tangent:

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

At $ \theta = \frac{\pi}{4} $, both the sine and cosine values are:

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

Therefore:

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

Find the point(s) where the derivative of cos(theta) equals zero on the filled out unit circle

To find where the derivative of $ \cos(\theta) $ equals zero, we first need to find the derivative:

$$ \frac{d}{d\theta} \cos(\theta) = -\sin(\theta) $$

Set the derivative to zero:

$$ -\sin(\theta) = 0 $$

Thus, we have:

$$ \sin(\theta) = 0 $$

The solutions to this equation on the unit circle are:

$$ \theta = 0, \pi, 2\pi $$

Therefore, the points on the unit circle are:

$$ (1, 0), (-1, 0), (1, 0) $$

Determine the coordinates of the points on the unit circle where the angle is pi/4

To determine the coordinates of the points on the unit circle where the angle is $ \frac{\pi}{4} $, we need to use trigonometric functions.

On the unit circle, the x-coordinate is given by $ \cos(\theta) $ and the y-coordinate is given by $ \sin(\theta) $, where $ \theta $ is the angle.

For $ \theta = \frac{\pi}{4} $:

$$ x = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

$$ y = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

Thus, the coordinates are:

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

Find the reference angle for a given angle of 345 degrees in the unit circle

To find the reference angle for $345^\circ$, note that it is in the fourth quadrant. The reference angle in the fourth quadrant is found by subtracting the given angle from $360^\circ$:

$$ 360^\circ – 345^\circ = 15^\circ $$

So, the reference angle for $345^\circ$ is:

$$ 15^\circ $$

Find the equation for a unit circle in the Cartesian plane

The equation for a unit circle centered at the origin in the Cartesian plane is:

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

This equation represents all points $(x, y)$ that are exactly one unit away from the origin.

Find the value of tan(π/4) using the unit circle

To find the value of $ \tan(\frac{\pi}{4}) $ using the unit circle:

On the unit circle, the coordinates for $ \frac{\pi}{4} $ are $ (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $.

Therefore:

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