Homework

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 sine and cosine of 150 degrees on the unit circle

To find the sine and cosine of $150^\circ$, we first identify its reference angle:

The reference angle for $150^\circ$ is:

$$180^\circ – 150^\circ = 30^\circ$$

The sine and cosine of $30^\circ$ are:

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

$$ \cos(30^\circ) = \frac{\sqrt{3}}{2} $$

Since $150^\circ$ is in the second quadrant, the sine is positive and the cosine is negative:

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

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

Find the equation of the inverse of the unit circle

The equation of the unit circle is:

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

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To find the inverse, we use the transformation:

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$$ z = \x0crac{1}{x + yi} $$

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where $ z = u + vi $ and $ x + yi = \x0crac{1}{u – vi} $.

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Therefore, the inverse relation in terms of $u$ and $v$ becomes:

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$$ u = \x0crac{x}{x^2 + y^2} = x $$

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$$ v = \x0crac{-y}{x^2 + y^2} = -y $$

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Thus, the equation of the inverse of the unit circle is:

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$$ u^2 + v^2 = 1 $$

Find the values of arcsin(1/2) using the unit circle

To find the values of $ \arcsin( \frac{1}{2} ) $ using the unit circle, we look for the angles $ \theta $ whose sine value is $ \frac{1}{2} $.

On the unit circle, the sine value is the y-coordinate. The angles with a y-coordinate of $ \frac{1}{2} $ are:

$$ \theta = \frac{\pi}{6} $$

or

$$ \theta = \frac{5\pi}{6} $$

So, the values of $ \arcsin( \frac{1}{2} ) $ are:

$$ \frac{\pi}{6} $ and $ \frac{5\pi}{6} $$

Determine the cosine of an angle on the unit circle at 45 degrees

To find the cosine of an angle at $45^{\circ}$ on the unit circle, recall that:

$$ \cos(45^{\circ}) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

Find the exact values of tan(theta) for theta on the unit circle at each 30-degree increment, and explain the symmetry properties of the tangent function on the unit circle

For each 30-degree increment ($ \theta $) on the unit circle, we have:

$ \tan(0^\circ) = 0 $

$ \tan(30^\circ) = \frac{1}{\sqrt{3}} $

$ \tan(60^\circ) = \sqrt{3} $

$ \tan(90^\circ) = \text{undefined} $

$ \tan(120^\circ) = -\sqrt{3} $

$ \tan(150^\circ) = -\frac{1}{\sqrt{3}} $

$ \tan(180^\circ) = 0 $

$ \tan(210^\circ) = \frac{1}{\sqrt{3}} $

$ \tan(240^\circ) = \sqrt{3} $

$ \tan(270^\circ) = \text{undefined} $

$ \tan(300^\circ) = -\sqrt{3} $

$ \tan(330^\circ) = -\frac{1}{\sqrt{3}} $

$ \tan(360^\circ) = 0 $

The tangent function is periodic with a period of $ 180^\circ $, hence $ \tan(\theta + 180^\circ) = \tan(\theta) $.

Find the exact values of trigonometric functions for given unit circle angles

Given the angle $ \theta = \frac{5\pi}{4} $, find the exact values of $ \sin(\theta) $, $ \cos(\theta) $, and $ \tan(\theta) $:

$$ \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} $$

$$ \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} $$

$$ \tan\left(\frac{5\pi}{4}\right) = 1 $$

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} $:

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

Find the angles at which sin(θ) = cos(θ)

To find the angles where $ \sin(\theta) = \cos(\theta) $, we know that:

$$ \sin(\theta) = \cos(\theta) $$

Dividing both sides by $ \cos(\theta) $, we get:

$$ \tan(\theta) = 1 $$

Thus, $ \theta $ must be an angle where the tangent is 1. We know that $ \tan(\theta) = 1 $ at:

$$ \theta = \frac{\pi}{4} + n\pi $$

where $ n $ is any integer. So, the angles are:

$$ \theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, … $$