Homework

Determine the quadrant of an angle of 45 degrees

To determine the quadrant of an angle of $45^{\circ}$, we need to look at the unit circle. Angles are measured counterclockwise from the positive x-axis.

Since $45^{\circ}$ is between $0^{\circ}$ and $90^{\circ}$, it lies in the first quadrant.

Therefore, the angle $45^{\circ}$ lies in the first quadrant.

Find the probability density function (pdf) for a uniform distribution on the unit circle

To find the probability density function (pdf) for a uniform distribution on the unit circle, we start by noting that the unit circle can be expressed in terms of its angular coordinate $\theta$, where $0 \leq \theta < 2\pi$.

Since the distribution is uniform, the probability density function (pdf) must be constant. The integral of the pdf over the entire circle must be 1:

$$ \int_0^{2\pi} f(\theta) \, d\theta = 1 $$

Let $f(\theta) = c$ be the constant pdf. Then:

$$ c \int_0^{2\pi} \, d\theta = 1 $$

Evaluating the integral gives:

$$ c \cdot 2\pi = 1 $$

Solving for $c$, we get:

$$ c = \frac{1}{2\pi} $$

Therefore, the pdf for a uniform distribution on the unit circle is:

$$ f(\theta) = \frac{1}{2\pi}, \quad 0 \leq \theta < 2\pi $$

Determine the reference angle for 5π/3 radians and express it in degrees and radians

To find the reference angle for $ \frac{5\pi}{3} $ radians, we need to determine its corresponding acute angle in the first quadrant.

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First, convert $ \frac{5\pi}{3} $ to degrees:

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$$ \frac{5\pi}{3} \times \frac{180^\circ}{\pi} = 300^\circ $$

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Since 300° is in the fourth quadrant, the reference angle is:

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$$ 360^\circ – 300^\circ = 60^\circ $$

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Convert 60° back to radians:

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$$ 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3} $$

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Therefore, the reference angle for $ \frac{5\pi}{3} $ radians is $ 60^\circ $ or $ \frac{\pi}{3} $ radians.

Given a unit circle, find the equation of a tangent line at a point (a,b) on the circle

To find the equation of the tangent line to the unit circle at the point $(a,b)$, recall that the unit circle is given by:

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

Since the radius at the point $(a,b)$ is perpendicular to the tangent, the slope of the radius is:

$$ m_r = \x0crac{b}{a} $$

Thus, the slope of the tangent line, being the negative reciprocal, is:

$$ m_t = -\x0crac{a}{b} $$

Using the point-slope form of a line, the tangent line equation is:

$$ y – b = -\x0crac{a}{b}(x – a) $$

Simplifying, we get:

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

Since $(a,b)$ lies on the unit circle, we know:

$$ a^2 + b^2 = 1 $$

So the tangent line equation simplifies to:

$$ y = -\x0crac{a}{b}x + \x0crac{1}{b} $$

Learn the values of sine and cosine on the unit circle

To learn the values of $ \sin $ and $ \cos $ on the unit circle, start with the key angles: $ 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} $. At these angles, memorize the coordinates $ (1, 0), (\frac{\sqrt{3}}{2}, \frac{1}{2}), (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}), (\frac{1}{2}, \frac{\sqrt{3}}{2}), (0, 1) $ respectively.

Find the value of arctan(1) and its corresponding point on the unit circle

To find the value of $ \arctan(1) $, we need to determine the angle whose tangent is 1. This angle is $ \frac{\pi}{4} $ radians or $ 45^{\circ} $.

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

Thus, the value of $ \arctan(1) $ is $ \frac{\pi}{4} $ and the corresponding point on the unit circle is $( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} )$.

Find the tan values for specific angles on the unit circle

To find the $\tan$ values for specific angles on the unit circle, we can use the unit circle properties. Let

Determine the values of tangent function in each quadrant on the unit circle

To determine the values of $\tan(\theta)$ in each quadrant on the unit circle, we use the properties of trigonometric functions:

In Quadrant I, where both sine and cosine are positive:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} > 0 $$

In Quadrant II, where sine is positive and cosine is negative:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} < 0 $$

In Quadrant III, where both sine and cosine are negative:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} > 0 $$

In Quadrant IV, where sine is negative and cosine is positive:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} < 0 $$

Find the cosine of π/3 using the unit circle

To find the cosine of $ \frac{\pi}{3} $ using the unit circle, follow these steps:

1. Locate the angle $ \frac{\pi}{3} $ on the unit circle.

2. The angle $ \frac{\pi}{3} $ corresponds to 60 degrees.

3. The coordinates of this angle on the unit circle are (1/2, \sqrt{3}/2).

4. The x-coordinate represents the cosine value.

Therefore, $ \cos(\frac{\pi}{3}) = \frac{1}{2} $.

Find the exact values of sin, cos, and tan for 7π/6 using the unit circle

To find the exact values of $ \sin, \cos, $ and $ \tan $ for $ \frac{7\pi}{6} $ using the unit circle, we need to determine the coordinates of the point corresponding to this angle.

Since $ \frac{7\pi}{6} $ is in the third quadrant, both sine and cosine values will be negative:

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

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

Now, using the identity $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $:

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