Homework

Find the value of tan(7π/6) and explain using the unit circle

To find the value of $ \tan(\frac{7\pi}{6}) $ using the unit circle:

1. Locate the angle $\frac{7\pi}{6}$ on the unit circle. This angle is in the third quadrant.

2. The reference angle for $\frac{7\pi}{6}$ is $\frac{\pi}{6}$.

3. In the third quadrant, both sine and cosine are negative. Knowing the coordinates for $\frac{\pi}{6}$ are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$:

The coordinates for $\frac{7\pi}{6}$ are $(-\frac{\sqrt{3}}{2}, -\frac{1}{2})$.

4. Finally, calculate the tangent value:

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

Finding the sine, cosine, and tangent of 45 degrees using the unit circle

To find the trigonometric functions of $45^\circ$ using the unit circle, note that $45^\circ$ corresponds to an angle in the first quadrant where both the x and y coordinates are equal.

Since the radius of the unit circle is 1, the coordinates of the point on the circle at $45^\circ$ are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

Thus:

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

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

$$ \tan(45^\circ) = 1 $$

Find the values of sin, cos, and tan for 45 degrees using the unit circle

To find the values of $ \sin $, $ \cos $, and $ \tan $ for $ 45^{\circ} $ using the unit circle, we start with:

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• $$ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} $$

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

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• $$ \tan(45^{\circ}) = 1 $$

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Determine the cosine of an angle corresponding to the point (1/2, sqrt(3)/2) on the unit circle

To determine the cosine of the angle corresponding to the point $ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $ on the unit circle, we must recognize the coordinates $(x, y)$ represent $(\cos(\theta), \sin(\theta))$.

In this case, the point is:

$$( \cos(\theta), \sin(\theta) ) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$$

Thus, the cosine of the angle is:

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

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

To find the coordinates of the point at an angle of $ \frac{\pi}{3} $ on the unit circle, we use the unit circle definition where the coordinates are given by $ (\cos\theta, \sin\theta) $.

For $ \theta = \frac{\pi}{3} $, we have:

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

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

So, the coordinates are:

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

Determine the sine and cosine values of an angle 𝜃 in radians on the unit circle given that 𝜃 = 5𝜋/4

Given $\theta = \frac{5\pi}{4}$, we determine the sine and cosine values by examining the unit circle.

The angle $\frac{5\pi}{4}$ is located in the third quadrant, where sine and cosine values are negative. Specifically:

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

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

Find the exact values of the inverse trigonometric functions on the unit circle

Consider the point $$P(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$$ on the unit circle. Determine the exact values for the following inverse trigonometric functions:

1. $$\arcsin(-\frac{\sqrt{3}}{2})$$

2. $$\arccos(-\frac{1}{2})$$

3. $$\arctan(\frac{\sqrt{3}}{3})$$

Answer:

1. $$\arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}$$

2. $$\arccos(-\frac{1}{2}) = \frac{2\pi}{3}$$

3. $$\arctan(\frac{\sqrt{3}}{3}) = \frac{\pi}{6}$$

Find the value of tan(θ) at θ = 3π/4 on the unit circle

To find the value of $ \tan(θ) $ at $ θ = \frac{3π}{4} $, we first identify the coordinates on the unit circle:

At $ θ = \frac{3π}{4} $, the coordinates are $ (-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $.

So, $ \tan(θ) $ is given by:

$$ \tan(θ) = \frac{y}{x} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

Therefore, $ \tan(θ) = -1 $.

Determine the values of sin, cos, and tan for an angle of 7π/6 on the unit circle

First, locate the angle $\frac{7\pi}{6}$ on the unit circle. This angle corresponds to 210 degrees.

The coordinates of the point on the unit circle at this angle are:

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

Thus:

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

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

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

Find the values of x for which cos(x) = -1/2 on the unit circle

To find the values of $\cos(x) = -\frac{1}{2}$ on the unit circle, we start by considering the unit circle where $\cos(\theta)$ is the x-coordinate of the point corresponding to the angle $\theta$:

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

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We know from trigonometric identities and the unit circle that:

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$$ \cos(120^\circ) = \cos\left( \frac{2\pi}{3} \right) = -\frac{1}{2} $$

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$$ \cos(240^\circ) = \cos\left( \frac{4\pi}{3} \right) = -\frac{1}{2} $$

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Therefore, the solutions in degrees are $120^\circ$ and $240^\circ$, and in radians they are:

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$$ x = \frac{2\pi}{3} + 2k\pi $$

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$$ x = \frac{4\pi}{3} + 2k\pi \quad \text{where } k \in \mathbb{Z} $$