Homework

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)} $:

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

Find the tan values of the unit circle at specific angles

To find the $\tan$ values of the unit circle at specific angles, we can use the fact that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$:

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

2. At $\theta = \frac{3\pi}{4}$, $$ \tan(\frac{3\pi}{4}) = \frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

3. At $\theta = \pi$, $$ \tan(\pi) = \frac{\sin(\pi)}{\cos(\pi)} = \frac{0}{-1} = 0 $$

Find the coordinates of the point on the unit circle corresponding to the angle 7π/6

To find the coordinates on the unit circle for the angle $\frac{7\pi}{6}$, we use the unit circle properties:

The unit circle coordinates $(x, y)$ for an angle $\theta$ are $(\cos(\theta), \sin(\theta))$.

For $\theta = \frac{7\pi}{6}$:

$$ x = \cos\left(\frac{7\pi}{6}\right) $$

$$ y = \sin\left(\frac{7\pi}{6}\right) $$

Using trigonometric identities:

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

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

Therefore, the coordinates are:

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

Find the values of sin(θ), cos(θ), and tan(θ) for θ = π/4 using the unit circle

To find the values of $\sin(\theta)$, $\cos(\theta)$, and $\tan(\theta)$ for $\theta = \frac{\pi}{4}$ using the unit circle, we use the following:

On the unit circle, at $\theta = \frac{\pi}{4}$, the coordinates are: $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.

So,

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

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

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