Homework

Determine the quadrant in which an angle lies given its sine and cosine values on the unit circle

Given that the sine and cosine values of an angle are both positive, the angle lies in the first quadrant.

Determine the coordinates of a point on the unit circle where the angle θ equals π/4

To determine the coordinates of a point on the unit circle where $ \theta $ equals $ \frac{\pi}{4} $, we use the unit circle equation:

$$ x^2 + y^2 = 1 $$

For $ \theta = \frac{\pi}{4} $, the coordinates are:

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

The values are:

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

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

To find the value of $ \sec(θ) $ at $ θ = \frac{\pi}{3} $ on the unit circle, we first find the cosine of the angle:

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

Then, since $ \sec(θ) $ is the reciprocal of $ \cos(θ) $:

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

Identify the coordinates of points on the unit circle for given angles

For the angle $ \theta = \frac{\pi}{6} $, the point on the unit circle is given by $ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) $.

Calculate these values:

$$ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $$

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

Therefore, the coordinates are:

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

Find the equation of a tangent to the unit circle at a given point

To find the equation of a tangent to the unit circle at the point $(a, b)$, we start by noting that the unit circle is defined by:

$$x^2 + y^2 = 1$$

The slope of the radius at $(a, b)$ is $ \x0crac{b}{a} $, so the slope of the tangent line, being perpendicular to the radius, is:

$$ -\x0crac{a}{b} $$

Using the point-slope form of a line, the equation of the tangent line can be written as:

$$ y – b = -\x0crac{a}{b}(x – a) $$

Simplifying, we get:

$$ bx + ay = 1 $$

Find the values of tan(θ) for θ in the unit circle at 0, π/4, π/3, and π/2

To determine the values of $ \tan(\theta) $ for $ \theta $ in the unit circle at $ 0 $, $ \frac{\pi}{4} $, $ \frac{\pi}{3} $, and $ \frac{\pi}{2} $, we evaluate the tangent function at these angles:

For $ \theta = 0 $:

$$ \tan(0) = 0 $$

For $ \theta = \frac{\pi}{4} $:

$$ \tan\left(\frac{\pi}{4}\right) = 1 $$

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

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

For $ \theta = \frac{\pi}{2} $:

$$ \tan\left(\frac{\pi}{2}\right) = \text{undefined} $$

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.