Unit Circle

Find the sine and cosine values at the angle pi/4

At the angle $ \frac{\pi}{4} $, the coordinates on the unit circle are:

$$ \left( \cos\left( \frac{\pi}{4} \right), \sin\left( \frac{\pi}{4} \right) \right) $$

Using the unit circle values:

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

Determine the quadrant of a given angle in radians on the unit circle

To determine the quadrant of an angle $ \theta $ in radians on the unit circle, follow these steps:

1. If $ \theta $ is greater than $ 2\pi $ or less than $ -2\pi $, reduce it by subtracting or adding multiples of $ 2\pi $ until it is within the range $ [0, 2\pi] $.

2. Check the reduced angle:

– If $ 0 \leq \theta < \frac{\pi}{2} $, the angle is in Quadrant I.

– If $ \frac{\pi}{2} \leq \theta < \pi $, the angle is in Quadrant II.

– If $ \pi \leq \theta < \frac{3\pi}{2} $, the angle is in Quadrant III.

– If $ \frac{3\pi}{2} \leq \theta < 2\pi $, the angle is in Quadrant IV.

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 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 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:

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