Unit Circle

Find the coordinates of the point on the unit circle corresponding to an angle of \( \frac{\pi}{4} \)

To find the coordinates of the point on the unit circle corresponding to an angle of $ \frac{\pi}{4} $, we use the unit circle definition:

The unit circle equation is: $$ x^2 + y^2 = 1 $$

At an angle of $ \frac{\pi}{4} $, both cosine and sine values are equal. Hence:

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

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

So, the coordinates are: $$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$

Determine the coordinates of a point on the unit circle at an angle of 11π/6

To determine the coordinates of a point on the unit circle at an angle of $$ \frac{11\pi}{6} $$, we use the unit circle properties.

The coordinates can be found using the cosine and sine of the angle:

$$ x = \cos \left( \frac{11\pi}{6} \right) $$

$$ y = \sin \left( \frac{11\pi}{6} \right) $$

Since $$ \frac{11\pi}{6} $$ is in the fourth quadrant, we know:

$$ \cos \left( \frac{11\pi}{6} \right) = \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} $$

$$ \sin \left( \frac{11\pi}{6} \right) = -\sin \left( \frac{\pi}{6} \right) = -\frac{1}{2} $$

So the coordinates are:

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

Solve for the exact values of all angles θ in the interval [0, 2π) that satisfy cos(θ) = -1/2

To find the exact values of all angles $ \theta $ in the interval $ [0, 2\pi) $ that satisfy $ \cos(\theta) = -\frac{1}{2} $, we use the unit circle. The cosine value of $ -\frac{1}{2} $ corresponds to angles in the second and third quadrants. The reference angle is $ \frac{\pi}{3} $.

The angles are:

• In the second quadrant: $ \pi – \frac{\pi}{3} = \frac{2\pi}{3} $
• In the third quadrant: $ \pi + \frac{\pi}{3} = \frac{4\pi}{3} $

Thus, the solutions are:

$$ \theta = \frac{2\pi}{3}, \frac{4\pi}{3} $$

Determine the coordinates of the point where the terminal side of an angle of 5π/3 radians intersects the unit circle, and identify its quadrant

The angle $ \frac{5\pi}{3} $ radians is equivalent to 300 degrees (since $ \frac{5\pi}{3} \times \frac{180}{\pi} = 300 $ degrees).

This angle places the terminal side in the fourth quadrant.

In the fourth quadrant, the coordinates on the unit circle corresponding to an angle of 300 degrees are:

$$ ( \cos(300\degree), \sin(300\degree) ) $$

Since $ \cos(300\degree) = \cos(-60\degree) = \frac{1}{2} $ and $ \sin(300\degree) = \sin(-60\degree) = -\frac{\sqrt{3}}{2} $, the coordinates are:

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

Thus, the terminal side intersects the unit circle at $ \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) $ in the fourth quadrant.

Find the values of tan(x) at different positions on the unit circle

To find the values of $ \tan(x) $ at different positions on the unit circle, we use the definition:

$$ \tan(x) = \frac{ \sin(x) }{ \cos(x) } $$

First, let

Calculate the integral of 1/(x + sqrt(x^2 – 1)) over the interval [-1, 1]

To calculate the integral:

$$ \int_{-1}^{1} \frac{1}{x + \sqrt{x^2 – 1}} dx $$

First, consider the substitution $ x = \cosh(t) $, which implies $ dx = \sinh(t) dt $.

When $ x = -1 $, $ t = i \pi $ and when $ x = 1 $, $ t = 0 $:

$$ \int_{i \pi}^{0} \frac{1}{\cosh(t) + \sinh(t)} \sinh(t) dt $$

Knowing that $ \cosh(t) + \sinh(t) = e^t $, the integral becomes:

$$ \int_{i \pi}^{0} \frac{\sinh(t)}{e^t} dt = \int_{i \pi}^{0} e^{-t} dt $$

Evaluating this gives:

$$ [ -e^{-t} ]_{i \pi}^{0} = -e^{0} + e^{-i \pi} = -1 + (-1) = -2 $$

Find the minimum value of cos(θ1+θ2+θ3) where θ1, θ2, θ3 are angles on the unit circle satisfying specific conditions

Let

Find the sine of a negative angle on the unit circle

On the unit circle, the sine of a negative angle $ \theta $ is given by:

$$ \sin(-\theta) = -\sin(\theta) $$

For example, if $ \theta = 30^{\circ} $, then:

$$ \sin(-30^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2} $$

Find the values of sin(θ), cos(θ), and tan(θ) at θ = π/4 on the unit circle

To find the values of $ \sin(\theta) $, $ \cos(\theta) $, and $ \tan(\theta) $ at $ \theta = \frac{\pi}{4} $ on the unit circle, we use the standard trigonometric values:

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

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

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

Determine the sine and cosine of an angle in the unit circle in the second quadrant

An angle $ \theta $ in the second quadrant of the unit circle ranges from $ 90^\circ $ to $ 180^\circ $ (or $ \frac{\pi}{2} $ to $ \pi $ radians). In this range, the sine of the angle is positive, and the cosine is negative.

For example, for $ \theta = 120^\circ $ (or $ \frac{2\pi}{3} $ radians):

$$ \sin(120^\circ) = \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} $$

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

Thus, the sine and cosine of an angle $ \theta $ in the second quadrant are:

$$ \sin(\theta) > 0 $$

$$ \cos(\theta) < 0 $$