Homework

What is the value of sin(15 degrees) using the unit circle?

To find the value of $ \sin(15^\circ) $ using the unit circle, we use the angle addition formula:

$$ \sin(a + b) = \sin(a) \cos(b) + \cos(a)\sin(b) $$

Here, let $ a = 45^\circ $ and $ b = -30^\circ $. Then,

$$ \sin(45^\circ – 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) $$

Substituting the values, we get:

$$ \sin(15^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} $$

Simplify the expression:

$$ \sin(15^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} $$

Combine the fractions:

$$ \sin(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} $$

Solve for the trigonometric value of tan(x) when cos(x) = 1/2 on the unit circle

To solve for $ \tan(x) $ given $ \cos(x) = \frac{1}{2} $ on the unit circle, we must first determine the corresponding $ \sin(x) $. On the unit circle:

$$ \cos(x) = \frac{1}{2} $$

We know that at $ x = \frac{\pi}{3} $ and $ x = -\frac{\pi}{3} $, $ \cos(x) = \frac{1}{2} $. Correspondingly, $ \sin(x) $ at these points are:

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

$$ \sin( – \frac{\pi}{3}) = – \frac{\sqrt{3}}{2} $$

Using the identity for tangent:

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

For $ x = \frac{\pi}{3} $:

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

For $ x = -\frac{\pi}{3} $:

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

Find the coordinates of the point on the unit circle that corresponds to an angle of pi/3

To find the coordinates of the point on the unit circle at an angle of $ \pi/3 $ , we use the fact that the unit circle

Complete the unit circle with the correct trigonometric values

To complete the unit circle, fill in the following values:

$$\text{1. } \sin(\frac{\pi}{6}) = \frac{1}{2}, \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$$

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

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

Find the coordinates on the unit circle at an angle of π/3

To find the coordinates on the unit circle at an angle of $ \frac{\pi}{3} $, we use the unit circle definition:

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

For $ \theta = \frac{\pi}{3} $:

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

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

The coordinates are:

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

Determine the values of theta that satisfy both sin(theta) = -1/2 and cos(theta) = -sqrt(3)/2

First, recognize that $ \sin(\theta) = -\frac{1}{2} $ in the third and fourth quadrants. The angles in these quadrants are $ \theta = \frac{7\pi}{6} $ and $ \theta = \frac{11\pi}{6} $.

Next, recognize that $ \cos(\theta) = -\frac{\sqrt{3}}{2} $ in the second and third quadrants. The angles in these quadrants are $ \theta = \frac{5\pi}{6} $ and $ \theta = \frac{7\pi}{6} $.

Therefore, the angle that satisfies both conditions is $ \theta = \frac{7\pi}{6} $.

Find the hypotenuse of a right triangle on the unit circle with one side equal to 1/√2

In a right triangle on the unit circle, the hypotenuse is always 1. If one side is $ \frac{1}{\sqrt{2}} $, the other side must also be $ \frac{1}{\sqrt{2}} $ to satisfy the Pythagorean theorem:

$$ a^2 + b^2 = c^2 $$

Here, $ a = \frac{1}{\sqrt{2}} $ and $ b = \frac{1}{\sqrt{2}} $, and the hypotenuse $ c = 1 $:

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

$$ \frac{1}{2} + \frac{1}{2} = 1 $$

Therefore, the hypotenuse is 1.

Determine the value of sin, cos, and tan for the angle θ = π/4 using the unit circle

Given the angle $\theta = \frac{\pi}{4}$, the corresponding coordinates on the unit circle are:

$$ (\cos(\theta), \sin(\theta)) = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) $$

Thus,

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

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

The tangent function is the ratio of sine to cosine:

$$ \tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = 1 $$

Find the coordinates of the point on the unit circle where the angle is 3π/4 radians

To find the coordinates of the point on the unit circle where the angle is $\frac{3\pi}{4}$ radians, we use the unit circle definition:

The coordinates are given by:

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

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

$$ \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} $$

$$ \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} $$

Therefore, the coordinates are:

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

Determine the exact values of the trigonometric functions for angle 7π/6

To determine the exact values of the trigonometric functions for angle $ \frac{7\pi}{6} $, we follow these steps:

1. Recognize that $ \frac{7\pi}{6} $ is in the third quadrant.
2. Calculate the reference angle:
$$ \pi – \frac{7\pi}{6} = \frac{\pi}{6} $$.

The sine and cosine values in the third quadrant are negative:

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

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

3. Calculate the tangent value:

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