Unit Circle

Find the sine and cosine of the angle π/3 using the unit circle

To find the sine and cosine of the angle $ \pi/3 $ using the unit circle, consider the angle that corresponds to $ \pi/3 $ radians (or 60 degrees).

In the unit circle, the coordinates of the point on the circumference corresponding to the angle $ \pi/3 $ are $ (\cos(\pi/3), \sin(\pi/3)) $.

For $ \pi/3 $:

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

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

Find the coordinates of the point where the terminal side of the angle intersects the unit circle at an angle of 5π/4 radians

To find the coordinates of the point where the terminal side of the angle intersects the unit circle at an angle of $\frac{5\pi}{4}$ radians, we use the unit circle properties.

The angle $\frac{5\pi}{4}$ radians is in the third quadrant where both sine and cosine values are negative.

The reference angle for $\frac{5\pi}{4}$ is $\frac{\pi}{4}$.

The coordinates on the unit circle for an angle of $\frac{\pi}{4}$ are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

Since we are in the third quadrant, we change the signs of both x and y coordinates:

Therefore, the coordinates are:

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

Find the coordinates of cos(π/3) on the unit circle

To find the coordinates of $ \cos(\frac{\pi}{3}) $ on the unit circle, we need to identify the coordinates associated with this angle.

On the unit circle, the angle $ \frac{\pi}{3} $ corresponds to the 60° position.

At this position, the coordinates are:

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

Determine the general solution for sin(x) = 1/2 within [0, 2π]

To determine the general solution for $ \sin(x) = \frac{1}{2} $ within the interval $ [0, 2\pi] $, we need to find all the angles $ x $ where the sine function yields $ \frac{1}{2} $ on the unit circle.

The sine function equals $ \frac{1}{2} $ at angles $ \frac{\pi}{6} $ and $ \frac{5\pi}{6} $ within the given interval.

Thus, the general solutions are:

$$ x = \frac{\pi}{6} + 2n\pi $$

and

$$ x = \frac{5\pi}{6} + 2n\pi $$

where $ n $ is any integer.

Determine the coordinates of points on the unit circle where the tangent line is horizontal

To find the coordinates on the unit circle where the tangent line is horizontal, we first recall that the unit circle is defined by the equation:

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

The slope of the tangent line to the circle at any point (x, y) is given by the derivative of y with respect to x. Differentiating implicitly, we get:

$$ 2x + 2y \x0crac{dy}{dx} = 0 $$

Solving for $\x0crac{dy}{dx}$, we find:

$$ \x0crac{dy}{dx} = -\x0crac{x}{y} $$

For the tangent line to be horizontal, the slope $\x0crac{dy}{dx}$ must be zero. This occurs when:

$$ -\x0crac{x}{y} = 0 $$

Thus, x must be zero. On the unit circle, the points with x = 0 are (0, 1) and (0, -1). Therefore, the coordinates are (0, 1) and (0, -1).

What are the coordinates of 3π/4 on the unit circle?

The coordinates of $ \frac{3\pi}{4} $ on the unit circle can be found using the unit circle definitions. The angle $ \frac{3\pi}{4} $ corresponds to $ 135^{\circ} $. At this angle, the coordinates are:

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

Find the sine and cosine of 7π/6 on the unit circle

To find the sine and cosine of $ \frac{7\pi}{6} $ on the unit circle, we first determine the reference angle. The reference angle for $ \frac{7\pi}{6} $ is $ \frac{\pi}{6} $.

The sine and cosine of $ \frac{7\pi}{6} $ correspond to the sine and cosine of $ \frac{\pi}{6} $ but with signs corresponding to the third quadrant.

From the unit circle, we know:

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

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

Since $ \frac{7\pi}{6} $ is in the third quadrant, where both sine and cosine are negative, we get:

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

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

Find the sine and cosine of 150 degrees on the unit circle

To find the sine and cosine of $150^\circ$, we first identify its reference angle:

The reference angle for $150^\circ$ is:

$$180^\circ – 150^\circ = 30^\circ$$

The sine and cosine of $30^\circ$ are:

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

$$ \cos(30^\circ) = \frac{\sqrt{3}}{2} $$

Since $150^\circ$ is in the second quadrant, the sine is positive and the cosine is negative:

$$ \sin(150^\circ) = \sin(30^\circ) = \frac{1}{2} $$

$$ \cos(150^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2} $$

Find the equation of the inverse of the unit circle

The equation of the unit circle is:

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

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To find the inverse, we use the transformation:

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$$ z = \x0crac{1}{x + yi} $$

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where $ z = u + vi $ and $ x + yi = \x0crac{1}{u – vi} $.

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Therefore, the inverse relation in terms of $u$ and $v$ becomes:

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$$ u = \x0crac{x}{x^2 + y^2} = x $$

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$$ v = \x0crac{-y}{x^2 + y^2} = -y $$

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Thus, the equation of the inverse of the unit circle is:

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$$ u^2 + v^2 = 1 $$

Find the values of arcsin(1/2) using the unit circle

To find the values of $ \arcsin( \frac{1}{2} ) $ using the unit circle, we look for the angles $ \theta $ whose sine value is $ \frac{1}{2} $.

On the unit circle, the sine value is the y-coordinate. The angles with a y-coordinate of $ \frac{1}{2} $ are:

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

or

$$ \theta = \frac{5\pi}{6} $$

So, the values of $ \arcsin( \frac{1}{2} ) $ are:

$$ \frac{\pi}{6} $ and $ \frac{5\pi}{6} $$