Unit Circle

Find the value of tan(θ) where θ is angle on the unit circle

Consider the angle $$\theta = \frac{\pi}{4}$$ on the unit circle.

We know that $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$.

Since $$\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$ and $$\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$, we get:

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

Therefore, $$\tan(\frac{\pi}{4}) = 1$$.

Find the value of sec(θ) on the unit circle where θ = 2π/3

Given $\theta = \frac{2\pi}{3}$, we need to find $\sec(\theta)$ on the unit circle.

First, we determine the coordinates of the point on the unit circle at $\theta = \frac{2\pi}{3}$.

The coordinates are $\left(\cos\left(\frac{2\pi}{3}\right), \sin\left(\frac{2\pi}{3}\right)\right)$. Since $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$, we have:

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

Thus, the value of $\sec\left(\frac{2\pi}{3}\right)$ is $-2$.

What is the value of \( \sin(\frac{\pi}{6}) \), \( \cos(\frac{\pi}{6}) \), and \( \tan(\frac{\pi}{6}) \)?

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

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

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

Find the value of cos(x) and sin(x) based on the unit circle

Given $x = \frac{5\pi}{4}$, we need to find $\cos(x)$ and $\sin(x)$ using the unit circle.

Step 1: Locate the angle $\frac{5\pi}{4}$ on the unit circle. This angle is in the third quadrant.

Step 2: Determine the reference angle. The reference angle for $\frac{5\pi}{4}$ is $\frac{\pi}{4}$.

Step 3: Recall the unit circle values for $\frac{\pi}{4}$. The coordinates are $(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ in the third quadrant.

Therefore, $\cos(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}}$ and $\sin(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}}$.

Find the angle θ in the interval [0, 2π) for which tan(θ) = -1

Consider the unit circle, where $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $.
For $ \tan(\theta) = -1 $, this implies that $ \sin(\theta) = -\cos(\theta) $.
Hence, $ \theta $ must be in the second or fourth quadrant, where sine and cosine have opposite signs.
This occurs at:

$$ \theta = \frac{3\pi}{4} $$

and

$$ \theta = \frac{7\pi}{4} $$

Therefore, the solutions to the equation $ \tan(\theta) = -1 $ in the interval $ [0, 2\pi) $ are:

$$ \theta = \frac{3\pi}{4} \text{ and } \frac{7\pi}{4} $$

In which quadrant of the unit circle is the angle 135 degrees located?

To determine the quadrant of the angle 135 degrees, we need to understand how the unit circle is divided:

– Quadrant I: $0^\circ$ to $90^\circ$

– Quadrant II: $90^\circ$ to $180^\circ$

– Quadrant III: $180^\circ$ to $270^\circ$

– Quadrant IV: $270^\circ$ to $360^\circ$

Since $135^\circ$ falls between $90^\circ$ and $180^\circ$, it is located in Quadrant II.

$$\text{Answer: Quadrant II}$$

Find the coordinates of a point on the unit circle with an angle of \( \frac{\pi}{4} \) radians

First, we need to recall that the unit circle is a circle with a radius of 1 centered at the origin.

For an angle of $ \frac{\pi}{4} $ radians, we can use the sine and cosine functions to find the coordinates.

The x-coordinate is given by $ \cos( \frac{\pi}{4} ) = \frac{\sqrt{2}}{2} $.

The y-coordinate is given by $ \sin( \frac{\pi}{4} ) = \frac{\sqrt{2}}{2} $.

Therefore, the coordinates of the point are $$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$.

Find the sine and cosine values of π/4 using the unit circle

To find the sine and cosine values of $\frac{\pi}{4}$ using the unit circle, follow these steps:

1. Identify the angle $\frac{\pi}{4}$ on the unit circle. This angle corresponds to 45 degrees.

2. For the angle $\frac{\pi}{4}$, both the x-coordinate (cosine) and the y-coordinate (sine) of the point on the unit circle are equal.

3. Using the symmetry of the unit circle, the sine and cosine values for $\frac{\pi}{4}$ are:

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

So, the sine and cosine values of $\frac{\pi}{4}$ are both $\frac{\sqrt{2}}{2}$.

Find the Value of Cosine on the Unit Circle

$$\text{Given the unit circle, we need to find the value of } \cos(\theta) \text{ where } \theta \text{ is an angle such that } 2\cos^2(\theta) + \cos(\theta) – 1 = 0.$$

$$\text{Step 1: Solve the quadratic equation} $$

$$2\cos^2(\theta) + \cos(\theta) – 1 = 0$$

$$\text{Using the quadratic formula } x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}, \text{ where } a = 2, b = 1, \text{ and } c = -1$$

$$\cos(\theta) = \frac{-1 \pm \sqrt{1^2 – 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$

$$\text{Thus, } \cos(\theta) = \frac{2}{4} = \frac{1}{2} \text{ or } \cos(\theta) = \frac{-4}{4} = -1.$$

$$\text{Therefore, the possible values of } \cos(\theta) \text{ are } \boxed{\frac{1}{2} \text{ and } -1}.$$

Find the cosine and sine of the angle 5π/6 using the unit circle

To find the cosine and sine of the angle $ \frac{5\pi}{6} $, we can use the unit circle. The angle $ \frac{5\pi}{6} $ is in the second quadrant, where the cosine is negative and the sine is positive.

First, find the reference angle:

$$ \text{Reference angle} = \pi – \frac{5\pi}{6} = \frac{\pi}{6} $$

For the angle $ \frac{\pi}{6} $, cosine and sine values are:

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

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

Since $ \frac{5\pi}{6} $ is in the second quadrant:

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

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