Unit Circle

Find the value of cos(π/3) using the unit circle on a graphing calculator

On the unit circle, the angle $\frac{\pi}{3}$ corresponds to 60 degrees. The coordinates of this point are $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. The x-coordinate of this point is $\cos(\frac{\pi}{3})$.

Therefore,

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

Find the exact values of sine and cosine for an angle of 225 degrees using the unit circle

To find the exact values of $ \sin $ and $ \cos $ for an angle of 225 degrees using the unit circle, we first convert the angle to radians:

$$ 225^\circ = 225 \times \frac{\pi}{180} = \frac{5\pi}{4} $$

The angle \( \frac{5\pi}{4} \) is in the third quadrant, where both sine and cosine are negative.

For an angle of \( \frac{5\pi}{4} \), we can reference the unit circle to see that:

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

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

Hence, the exact values of sine and cosine for 225 degrees are:

$$ \sin(225^\circ) = -\frac{\sqrt{2}}{2} $$

$$ \cos(225^\circ) = -\frac{\sqrt{2}}{2} $$

Find the sine and cosine of \( \frac{\pi}{4} \) using the unit circle

To find the sine and cosine of $ \frac{\pi}{4} $ using the unit circle:

On the unit circle, the angle $$ \frac{\pi}{4} $$ corresponds to the coordinates $$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$.

Therefore,

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

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

Determine the coordinates of a point on the unit circle at an angle of \( \frac{\pi}{4} \)

To find the coordinates of a point on the unit circle at an angle of \( \frac{\pi}{4} \), we use the unit circle definition:

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The unit circle is defined as all points (x, y) such that:

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

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For an angle \( \theta \), the coordinates are given by:

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

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At \( \theta = \frac{\pi}{4} \):

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

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

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So, the coordinates are:

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

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:

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