Math

Find the tangent of angle θ on a unit circle

To find the tangent of the angle $ \theta $ on a unit circle, one must understand that the tangent of an angle is defined as the ratio of the sine to the cosine of that angle:

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

For example, if $ \theta = \frac{\pi}{4} $:

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

So:

$$ \tan(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

Find the values of sin, cos, and tan for angles that satisfy the equation 2sin(x)cos(x) = 1

First, recognize that $$2\sin(x)\cos(x) = \sin(2x)$$. Thus, the equation becomes:

$$\sin(2x) = 1$$

The solution for $$\sin(2x) = 1$$ occurs at:

$$2x = \frac{\pi}{2} + 2k\pi$$, where $$k$$ is any integer.

Thus:

$$x = \frac{\pi}{4} + k\pi$$

For $$k = 0$$:

$$x = \frac{\pi}{4}$$

Then:

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

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

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

For $$k = 1$$:

$$x = \frac{5\pi}{4}$$

Then:

$$\sin(x) = \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

$$\cos(x) = \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

$$\tan(x) = \tan\left(\frac{5\pi}{4}\right) = 1$$

Find the coordinates of a point on the negative unit circle given a specific angle

To find the coordinates of a point on the negative unit circle given a specific angle $ \theta $, we use the equation of the unit circle:

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

The coordinates can be found using parametric equations:

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

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

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

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

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

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

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

Determine the points on the negative unit circle where the tangent line is vertical

The negative unit circle is described by the equation:

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

To find where the tangent line is vertical, we need to find the points where the derivative of $ y $ with respect to $ x $ is undefined. First, implicitly differentiate the equation:

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

Solving for $ \x0crac{dy}{dx} $:

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

The derivative is undefined when $ y = 0 $. Substituting $ y = 0 $ into the original equation:

$$ x^2 = -1 $$

This has no real solutions. Therefore, there are no points on the negative unit circle where the tangent line is vertical.

Prove that the sum of the squares of the sine and cosine functions on the unit circle equals 1

On the unit circle, any point is represented as $(\cos(\theta), \sin(\theta))$, where $\theta$ is the angle formed with the positive x-axis.

According to the Pythagorean theorem, the equation of the unit circle is:

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

Substituting the coordinates:

$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$

Therefore, the sum of the squares of the sine and cosine functions on the unit circle equals 1.

Find the coordinates of the point on the unit circle for angle π/3

For the angle $ \frac{\pi}{3} $ on the unit circle, the coordinates are found using the sine and cosine functions.

The x-coordinate is:

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

The y-coordinate is:

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

Thus, the coordinates are:

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

Determine the value of cos(7π/6) using the unit circle

To determine the value of $ \cos\left(\frac{7\pi}{6}\right) $ using the unit circle, we need to locate the angle $ \frac{7\pi}{6} $ in radians. This angle is in the third quadrant.

In the third quadrant, the cosine function is negative. The reference angle for $ \frac{7\pi}{6} $ is $ \frac{\pi}{6} $, whose cosine value is $ \frac{\sqrt{3}}{2} $.

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

Determine the sine and cosine values for an angle of 5π/6 radians on the unit circle

To find the sine and cosine of $ \frac{5\pi}{6} $ on the unit circle, we use the reference angle and the fact that it lies in Quadrant II:

The reference angle for $ \frac{5\pi}{6} $ is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.

In Quadrant II, sine is positive and cosine is negative. Therefore:

$$ \sin\left( \frac{5\pi}{6} \right) = \sin\left( \frac{\pi}{6} \right) = \frac{1}{2} $$

$$ \cos\left( \frac{5\pi}{6} \right) = -\cos\left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2} $$

Express the coordinates of key points on the unit circle in terms of trigonometric functions

To express the coordinates of key points on the unit circle in terms of trigonometric functions, remember that each point on the unit circle corresponds to an angle $\theta$ and can be written as $(\cos(\theta), \sin(\theta))$. For example:

For $\theta = 0$: $$\cos(0) = 1, \sin(0) = 0$$ Hence, the coordinates are $(1, 0)$.

For $\theta = \frac{\pi}{2}$: $$\cos\left(\frac{\pi}{2}\right) = 0, \sin\left(\frac{\pi}{2}\right) = 1$$ Hence, the coordinates are $(0, 1)$.

For $\theta = \pi$: $$\cos(\pi) = -1, \sin(\pi) = 0$$ Hence, the coordinates are $(-1, 0)$.

For $\theta = \frac{3\pi}{2}$: $$\cos\left(\frac{3\pi}{2}\right) = 0, \sin\left(\frac{3\pi}{2}\right) = -1$$ Hence, the coordinates are $(0, -1)$.

Find the value of arcsin(x) for x = sqrt(3)/2 on the unit circle

To find the value of $ \arcsin(x) $ for $ x = \sqrt{3}/2 $ on the unit circle, we need to determine the angle $ \theta $ such that $ \sin(\theta) = \sqrt{3}/2 $ and $ \theta $ lies in the range $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.

The angle $ \theta $ corresponding to $ \sin(\theta) = \sqrt{3}/2 $ is $ \frac{\pi}{3} $.

Hence, $ \arcsin(\sqrt{3}/2) = \frac{\pi}{3} $.