Unit Circle

Find the value of arcsin(1/2)

To find the value of $ \arcsin(\frac{1}{2}) $, we need to determine the angle $ \theta $ whose sine is $ \frac{1}{2} $.

From the unit circle, we know:

$$ \sin(\theta) = \frac{1}{2} $$

The angle $ \theta $ that satisfies this in the range $ [-\frac{\pi}{2}, \frac{\pi}{2}] $ is:

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

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

Find the coordinates on the unit circle corresponding to an angle theta

To find the coordinates on the unit circle corresponding to an angle $ \theta $ , we use the parametric equations of the unit circle:

$$ x = \cos(\theta) $$

$$ y = \sin(\theta) $$

Thus, the coordinates are given by:

$$ (x, y) = (\cos(\theta), \sin(\theta)) $$

How to find the reference angle for an angle not located on the unit circle

To find the reference angle for an angle not located on the unit circle, we first need to understand that the reference angle is the smallest angle between the given angle and the x-axis. Let\

What is the value of cos(-π/3) using the unit circle?

To find the value of $\cos(-\frac{\pi}{3})$ using the unit circle, first recognize that the cosine function is an even function. This means that:

$$\cos(-x) = \cos(x)$$

Therefore:

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

From the unit circle, we know that:

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

Thus:

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

Find the coordinates of the point on the unit circle at which the angle is 7π/6

To find the coordinates of the point on the unit circle at which the angle is $ \frac{7\pi}{6} $, we use the following:

The unit circle has the equation:

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

The coordinates of a point on the unit circle are given by:

$$ (\cos(\theta), \sin(\theta)) $$

For $ \theta = \frac{7\pi}{6} $:

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

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

Therefore, the coordinates are:

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

Find the exact coordinates of the point(s) on the unit circle where the tangent line is vertical

The equation of the unit circle is given by:

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

We find the tangent line to be vertical when the derivative is undefined. Thus, we need to find the points where $ \x0crac{dy}{dx} $ is undefined.

Implicitly differentiate the unit circle equation with respect to $ x $:

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

Simplify and solve for $ \x0crac{dy}{dx} $:

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

The derivative is undefined when $ y = 0 $. Thus, we solve for $ x $:

When $ y = 0 $, substituting back into the original equation:

$$ x^2 = 1 $$

So, $ x = 1 $ or $ x = -1 $.

Therefore, the points are:

$(1,0)$ and $(-1,0)$.

Find the sine and cosine of the angle when the point is (1/2, √3/2) on the unit circle

The coordinates \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \) on the unit circle represent the cosine and sine of an angle:

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

The angle in radians corresponding to these values is:

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

Evaluate the integral of sin(x) * cos(x) around the unit circle

To evaluate the integral of $ \sin(x) * \cos(x) $ around the unit circle, we can use the double-angle identity:

$$ \sin(x) \cos(x) = \frac{1}{2} \sin(2x) $$

Now, we need to integrate from $ 0 $ to $ 2\pi $:

$$ \int_{0}^{2\pi} \sin(x) \cos(x) \, dx = \int_{0}^{2\pi} \frac{1}{2} \sin(2x) \, dx $$

Let $ u = 2x $, hence $ du = 2 \, dx $ and $ dx = \frac{1}{2} du $:

$$ \int_{0}^{2\pi} \frac{1}{2} \sin(2x) \, dx = \frac{1}{2} \int_{0}^{4\pi} \sin(u) \frac{1}{2} \, du $$

Combining constants:

$$ \frac{1}{4} \int_{0}^{4\pi} \sin(u) \, du $$

The integral of $ \sin(u) $ over one period is zero, and here we have two periods:

$$ \frac{1}{4} \left(0\right) = 0 $$

The integral evaluates to $ 0 $.

Find the terminal point on the unit circle for an angle of pi/6 radians

To find the terminal point on the unit circle for an angle of $ \frac{\pi}{6} $ radians, we use the unit circle definition:

The coordinates are given by $ ( \cos( \theta ), \sin( \theta ) ) $.

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

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

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

So, the terminal point is:

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

Identify the sine and cosine values for the angle π/4 on the unit circle

To find the sine and cosine values for the angle $ \frac{\pi}{4} $ on the unit circle, we use the definitions of sine and cosine:

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

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