Unit Circle

Find the angle $ \theta $ on the unit circle where the following conditions are met: $ \sin(\theta) = -\frac{1}{2} $ and $ \cos(\theta) = -\frac{\sqrt{3}}{2} $

To find the angle $ \theta $ on the unit circle where $ \sin(\theta) = -\frac{1}{2} $ and $ \cos(\theta) = -\frac{\sqrt{3}}{2} $, we need to identify the corresponding angles in degrees.

First, note that $ \sin(\theta) = -\frac{1}{2} $ occurs at:

$$ \theta = 210^\circ, 330^\circ $$

Next, note that $ \cos(\theta) = -\frac{\sqrt{3}}{2} $ occurs at:

$$ \theta = 150^\circ, 210^\circ $$

The common angle is:

$$ \theta = 210^\circ $$

Find the value of csc(θ) when θ = π/6 on the unit circle

To find the value of $csc(θ)$ when $θ = \frac{π}{6}$ on the unit circle, we first find the sine of $θ$:

$$ \sin(\frac{π}{6}) = \frac{1}{2} $$

Since $csc(θ) = \frac{1}{\sin(θ)}$, we have:

$$ csc(θ) = \frac{1}{\sin(\frac{π}{6})} = \frac{1}{\frac{1}{2}} = 2 $$

Find the exact value of cos(pi/9) using the unit circle and trigonometric identities

To find the exact value of $ \cos(\frac{\pi}{9}) $, we can use the triple-angle identity for cosine:

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$$ \cos(3\theta) = 4\cos^3(\theta) – 3\cos(\theta) $$

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Letting $ \theta = \frac{\pi}{9} $, we get:

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$$ \cos(\frac{\pi}{3}) = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9}) $$

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Since $ \cos(\frac{\pi}{3}) = \frac{1}{2} $, substituting in we have:

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$$ \frac{1}{2} = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9}) $$

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Let $ x = \cos(\frac{\pi}{9}) $, then the equation becomes:

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$$ \frac{1}{2} = 4x^3 – 3x $$

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Multiplying through by 2 to clear the fraction:

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$$ 1 = 8x^3 – 6x $$

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This is a cubic equation that can be solved for $ x = \cos(\frac{\pi}{9}) $ using numerical methods or by recognizing that:

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$$ \cos(\frac{\pi}{9}) = \frac{\sqrt{6} + \sqrt{2}}{4} $$

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:

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