Math

Find the coordinates of point on the unit circle

Given a point $(x, y)$ on the unit circle, we know that the equation of the circle is:

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

If $ x = \frac{1}{2} $, then we can find $ y $ by solving:

$$ (\frac{1}{2})^2 + y^2 = 1 $$

$$ \frac{1}{4} + y^2 = 1 $$

Solving for $ y $:

$$ y^2 = 1 – \frac{1}{4} $$

$$ y^2 = \frac{3}{4} $$

$$ y = \pm \frac{\sqrt{3}}{2} $$

So the coordinates are:

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

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.