Unit Circle

Evaluate the integral of sec(x) along the unit circle

To evaluate the integral of $ \sec(x) $ along the unit circle, we consider the parametrization of the unit circle. The unit circle can be parametrized as $ x = \cos(\theta) $ and $ y = \sin(\theta) $, where $ \theta $ ranges from $ 0 $ to $ 2\pi $.

The integral to evaluate becomes:

$$ \int_0^{2\pi} \sec(\cos(\theta)) \frac{d\theta}{d \theta} \ d\theta $$

We need to express $ \sec(\cos(\theta)) $ in terms of $ \theta $. However, since $ \sec(x) $ is not straightforward to integrate on the unit circle, it is more practical to use a different approach, often involving complex analysis or residue theorem.

Prove that sin(π/6) using the unit circle

To prove that $ \sin(\frac{\pi}{6}) $ using the unit circle, we start by locating the angle $ \frac{\pi}{6} $ on the unit circle.

The angle $ \frac{\pi}{6} $ corresponds to 30 degrees.

Using the unit circle, we see that the coordinates for this angle are $ (\frac{\sqrt{3}}{2}, \frac{1}{2}) $.

Since the sine of an angle is the y-coordinate on the unit circle, we have:

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

Determine the values of cos(θ) and sin(θ) using the unit circle when 0 ≤ θ ≤ 2π and θ is a solution to the equation tan(θ) = √3

The equation $\tan(\theta) = \sqrt{3}$ implies that:

$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \sqrt{3}$$

This happens at $\theta = \frac{\pi}{3}$ and $\theta = \frac{4\pi}{3}$ within the interval $0 ≤ \theta ≤ 2\pi$.

At $\theta = \frac{\pi}{3}$:

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

At $\theta = \frac{4\pi}{3}$:

$$\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}, \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$

Thus, the values are:

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

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

Find the value of angle θ where cos(θ) = -1/2 on the unit circle

The cosine function represents the x-coordinate on the unit circle. Thus, finding $ \cos(\theta) = -\frac{1}{2} $ involves finding the angles where the x-coordinate is -1/2. On the unit circle, this occurs at:

$$ \theta = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{4\pi}{3} + 2k\pi $$

for any integer $ k $.

Find the angle in radians and degrees for the point (-1/2, -√3/2) on the unit circle

We need to find the angle corresponding to the point $ \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) $ on the unit circle. This point lies in the third quadrant where both sine and cosine are negative. The reference angle is given by:

$$ \text{Reference angle} = \arccos\left( \frac{1}{2} \right) = \frac{\pi}{3} $$

Since the point is in the third quadrant, the angle in radians is:

$$ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $$

To convert this to degrees:

$$ \theta = \frac{4\pi}{3} \times \frac{180}{\pi} = 240^{\circ} $$

Hence, the angle is $ \frac{4\pi}{3} $ radians or $ 240^{\circ} $.

Find the value of arcsin(1/2) in radians using the unit circle

To find the value of $ \arcsin(\frac{1}{2}) $, consider the unit circle and the definition of arcsin. The arcsin function outputs the angle whose sine is the given value within the range $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $.

For $ \arcsin(\frac{1}{2}) $, we need to find the angle $ \theta $ such that $ \sin(\theta) = \frac{1}{2} $. On the unit circle, $ \sin(30^{\circ}) = \frac{1}{2} $ or equivalently, in radians:

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

Thus, the value of $ \arcsin(\frac{1}{2}) $ is:

$$ \arcsin(\frac{1}{2}) = \frac{\pi}{6} $$

Find the value of cosine at $\theta = \frac{3\pi}{4}$

The unit circle helps us locate the angle $\theta = \frac{3\pi}{4}$ which lies in the second quadrant. The reference angle for $\theta = \frac{3\pi}{4}$ is:

$$\pi – \frac{3\pi}{4} = \frac{\pi}{4}$$

In the second quadrant, the cosine of an angle is negative:

$$\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right)$$

Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we have:

$$\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

Find the coordinates of the point where the angle π/4 intersects the unit circle

To find the coordinates of the point where the angle $ \frac{\pi}{4} $ intersects the unit circle, we use the unit circle definition. The unit circle has a radius of 1, and the coordinates of the points on the circle are given by the cosine and sine of the angle.

For the angle $ \frac{\pi}{4} $, we have:

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

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

Therefore, the coordinates are:

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

How to find the reference angle for any angle not on the unit circle

To find the reference angle for an angle θ not on the unit circle, you must first locate the angle in the appropriate quadrant. The reference angle is then the smallest angle between the terminal side of θ and the x-axis. Here are the steps:

1. If θ is in the first quadrant, the reference angle is θ itself:

$$ θ_{ref} = θ $$

2. If θ is in the second quadrant, the reference angle is:

$$ θ_{ref} = 180° – θ $$

3. If θ is in the third quadrant, the reference angle is:

$$ θ_{ref} = θ – 180° $$

4. If θ is in the fourth quadrant, the reference angle is:

$$ θ_{ref} = 360° – θ $$

Given that cos(θ) = -1/2, find the general solutions for θ in the unit circle

To solve for $ θ $ such that $ \cos(θ) = -\frac{1}{2} $, we need to find all angles in the unit circle where the cosine value is $ -\frac{1}{2} $. The cosine function is negative in the second and third quadrants.

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The reference angle for $ \cos(θ) = \frac{1}{2} $ is $ \frac{\pi}{3} $. Therefore, the general solutions in the second and third quadrants are:

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$$ θ = \pi – \frac{\pi}{3} + 2k\pi = \frac{2\pi}{3} + 2k\pi $$

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and

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$$ θ = \pi + \frac{\pi}{3} + 2k\pi = \frac{4\pi}{3} + 2k\pi $$

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where $ k $ is any integer.