Unit Circle

Explore the unit circle and its relationship to angles, radians, trigonometric ratios, and coordinates in the coordinate plane.

Determine the Location of -π/2 on a Unit Circle

Determine the Location of -π/2 on a Unit Circle

To determine the location of $-\pi/2$ on a unit circle, we follow these steps:

1. Understand that the unit circle is a circle with a radius of 1 centered at the origin (0,0) in the coordinate plane.

2. The angle $-\pi/2$ is measured in radians and indicates a rotation of 90 degrees in the clockwise direction from the positive x-axis.

3. On the unit circle, $-\pi/2$ radians corresponds to the point where the angle terminates. Moving 90 degrees clockwise from the positive x-axis places the terminal side of the angle along the negative y-axis.

Therefore, the coordinates of the point corresponding to $-\pi/2$ are:

$$(-\pi/2) = (0, -1)$$

Thus, the point on the unit circle corresponding to the angle $-\pi/2$ is (0, -1).

Find the sine, cosine, and tangent values for the angle $\theta = \frac{5\pi}{6}$ using the unit circle

Find the sine, cosine, and tangent values for the angle $\theta = \frac{5\pi}{6}$ using the unit circle

For the angle $\theta = \frac{5\pi}{6}$:

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

In the second quadrant, sine is positive, cosine is negative, and tangent is negative.

Thus, the values are:

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

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

$\tan(\frac{5\pi}{6}) = \frac{\sin(\frac{5\pi}{6})}{\cos(\frac{5\pi}{6})} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

Ways to Memorize the Unit Circle

Ways to Memorize the Unit Circle

$$Ways to Memorize the Unit Circle$$

Explanation with Examples:

The unit circle is a circle with a radius of 1 centered at the origin (0,0) in the coordinate plane. To memorize the unit circle, follow these steps.

1. Know the Key Angles:

Memorize the common angles in radians: 0, $$\frac{\pi}{6}$$, $$\frac{\pi}{4}$$, $$\frac{\pi}{3}$$, $$\frac{\pi}{2}$$, $$\pi$$, $$\frac{3\pi}{2}$$, and $$2\pi$$.

2. Memorize the Coordinates:

For each angle, memorize the coordinates on the unit circle.

For instance:

$$\text{At }\theta = 0\text{ or }2\pi,$$

$$(cos(\theta), sin(\theta)) = (1, 0)$$

$$\text{At }\theta = \frac{\pi}{2},$$

$$(cos(\theta), sin(\theta)) = (0, 1)$$

$$\text{At }\theta = \pi,$$

$$(cos(\theta), sin(\theta)) = (-1, 0)$$

$$\text{At }\theta = \frac{3\pi}{2},$$

$$(cos(\theta), sin(\theta)) = (0, -1)$$

3. Use Mnemonics:

Use mnemonic devices to remember the coordinates, such as the phrase ‘All Students Take Calculus’ to remember the signs of the coordinates in each quadrant.

Find the coordinates of a point on the unit circle corresponding to a given angle

Find the coordinates of a point on the unit circle corresponding to a given angle

Given an angle of \( \theta = 45^{\circ} \). To find the coordinates of the point on the unit circle:

The coordinates of any point on the unit circle can be found using the formulas:

\[ x = \cos(\theta) \]

\[ y = \sin(\theta) \]

Using \( \theta = 45^{\circ} \):

\[ x = \cos(45^{\circ}) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \]

\[ y = \sin(45^{\circ}) = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \]

The coordinates are \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Find the value of tan(θ) on the unit circle where θ is π/4

Find the value of tan(θ) on the unit circle where θ is π/4

On the unit circle, the coordinates for $\theta = \frac{\pi}{4}$ are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

Therefore, $\tan(\frac{\pi}{4})$ is calculated as:

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

If point P on the unit circle is flipped over the y-axis, what will be the coordinates of point P if it initially lies on the point (sqrt(3)/2, 1/2)?

If point P on the unit circle is flipped over the y-axis, what will be the coordinates of point P if it initially lies on the point (sqrt(3)/2, 1/2)?

Initial coordinates of point $P$ are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$. When flipped over the $y$-axis, the x-coordinate becomes its negative value while the y-coordinate remains the same. Therefore, the new coordinates of point $P$ are:

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

Find the coordinates of a point on the unit circle at 45 degrees

Find the coordinates of a point on the unit circle at 45 degrees

The unit circle is a circle with a radius of 1 centered at the origin. To find the coordinates of a point at $45^{\circ}$, we use the trigonometric functions sine and cosine.

For $\theta = 45^{\circ}$:

$$\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$$

$$\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$$

Therefore, the coordinates of the point are:

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

Find the coordinates of the points on the unit circle where the angle formed with the positive x-axis is such that the cosine of the angle equals -3/5 Additionally, find the corresponding sine value

Find the coordinates of the points on the unit circle where the angle formed with the positive x-axis is such that the cosine of the angle equals -3/5 Additionally, find the corresponding sine value

To solve this problem, we start with the unit circle equation:

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

Given that $\cos(\theta) = \frac{-3}{5}$, we know the x-coordinate is $\frac{-3}{5}$. Let’s find the y-coordinate (sine value).

Substituting $\cos(\theta)$ in the unit circle equation:

$$\left(\frac{-3}{5}\right)^2 + y^2 = 1$$

$$\frac{9}{25} + y^2 = 1$$

Solving for $y^2$:

$$y^2 = 1 – \frac{9}{25}$$

$$y^2 = \frac{25}{25} – \frac{9}{25}$$

$$y^2 = \frac{16}{25}$$

Thus, $y = \pm \frac{4}{5}$.

The coordinates on the unit circle are:

$$\left( \frac{-3}{5}, \frac{4}{5} \right) \text{ and } \left( \frac{-3}{5}, \frac{-4}{5} \right)$$

Hence, the coordinates are $\left( \frac{-3}{5}, \frac{4}{5} \right) \text{ and } \left( \frac{-3}{5}, \frac{-4}{5} \right)$, and the corresponding sine values are $\frac{4}{5}$ and $\frac{-4}{5}$.

Find the Cartesian coordinates of a point on the unit circle where the angle is 135 degrees

Find the Cartesian coordinates of a point on the unit circle where the angle is 135 degrees

To find the Cartesian coordinates of a point on the unit circle where the angle is $135^{\circ}$, we use the unit circle equation:

$$x = \cos(135^{\circ})$$

$$y = \sin(135^{\circ})$$

First, we calculate the cosine and sine of $135^{\circ}$:

$$\cos(135^{\circ}) = -\frac{\sqrt{2}}{2}$$

$$\sin(135^{\circ}) = \frac{\sqrt{2}}{2}$$

So, the Cartesian coordinates are:

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

Find the exact value of sin(2θ) and cos(2θ) given the point on the unit circle and the quadrant

Find the exact value of sin(2θ) and cos(2θ) given the point on the unit circle and the quadrant

Given the point $ P = \left( -\frac{3}{5}, -\frac{4}{5} \right) $ on the unit circle, find the exact values of $ \sin(2\theta) $ and $ \cos(2\theta) $. The point $ P $ is in Quadrant III.

To find $ \theta $, we use the definitions of sine and cosine on the unit circle:

$$ \sin(\theta) = y = -\frac{4}{5} $$

$$ \cos(\theta) = x = -\frac{3}{5} $$

Using the double angle formulas:

$$ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) $$

$$ \cos(2\theta) = \cos^2(\theta) – \sin^2(\theta) $$

Substituting the values:

$$ \sin(2\theta) = 2 \left( -\frac{4}{5} \right) \left( -\frac{3}{5} \right) = 2 \times \frac{12}{25} = \frac{24}{25} $$

$$ \cos(2\theta) = \left( -\frac{3}{5} \right)^2 – \left( -\frac{4}{5} \right)^2 = \frac{9}{25} – \frac{16}{25} = -\frac{7}{25} $$

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