Homework

Fill in the missing value on the unit circle: What is the value of sin(θ) when cos(θ) = -1/2?

Given the unit circle equation:

$$\cos^2(θ) + \sin^2(θ) = 1$$

Substitute $\cos(θ) = -\frac{1}{2}$:

$$\left(-\frac{1}{2}\right)^2 + \sin^2(θ) = 1$$

Simplify:

$$\frac{1}{4} + \sin^2(θ) = 1$$

Subtract $\frac{1}{4}$ from both sides:

$$\sin^2(θ) = 1 – \frac{1}{4}$$

$$\sin^2(θ) = \frac{3}{4}$$

Taking the square root of both sides:

$$\sin(θ) = \pm \sqrt{\frac{3}{4}}$$

$$\sin(θ) = \pm \frac{\sqrt{3}}{2}$$

So, the value of $\sin(θ)$ when $\cos(θ) = -\frac{1}{2}$ is $\pm \frac{\sqrt{3}}{2}$.

Find the values of sin, cos, and tan for 45 degrees on the unit circle

Given $\theta = 45^\circ$, we need to find the values of $\sin(45^\circ)$, $\cos(45^\circ)$, and $\tan(45^\circ)$.

From the unit circle, we know that $\sin(45^\circ) = \frac{\sqrt{2}}{2}$ and $\cos(45^\circ) = \frac{\sqrt{2}}{2}$.

Using the identity $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$, we get:

$$\tan(45^\circ) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$$

Therefore, $\sin(45^\circ) = \frac{\sqrt{2}}{2}$, $\cos(45^\circ) = \frac{\sqrt{2}}{2}$, and $\tan(45^\circ) = 1$.

Find the sine and cosine values for a given angle on the unit circle

Suppose we need to find the sine and cosine values for the angle $ \frac{\pi}{4} $ on the unit circle.

Step 1: Recognize the angle $ \frac{\pi}{4} $ on the unit circle.

Step 2: Recall that $ \frac{\pi}{4} $ corresponds to 45 degrees.

Step 3: Know that the coordinates at 45 degrees (or $ \frac{\pi}{4} $ radians) on the unit circle are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

Step 4: Therefore, $ \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $ and $ \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $.

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

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

Determine the sin value from the unit circle and verify identities

Given the point $P(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$ on the unit circle, determine $\sin(\theta)$ and verify the identity $\sin^2(\theta) + \cos^2(\theta) = 1$:

1. Identify the coordinates of point $P$ as $(\cos(\theta), \sin(\theta))$.

2. From $P(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$, we have $\cos(\theta) = -\frac{1}{2}$ and $\sin(\theta) = -\frac{\sqrt{3}}{2}$.

3. Verify the identity:

$$\sin^2(\theta) + \cos^2(\theta) = \left(-\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2$$

$$= \frac{3}{4} + \frac{1}{4} = 1$$

The identity is verified as true.

Find the tangent for every angle on the unit circle

To find the tangent for every angle \(\theta\) on the unit circle, we use the definition of tangent, which is the ratio of the sine of the angle to the cosine of the angle.

The tangent function is given by:

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

To illustrate, let’s take an angle \(\theta = \frac{\pi}{4}\):

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

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

Thus,

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

Find the coordinates of the points where the line y = 1/2 intersects the unit circle

To find the intersection points of the line $y = \frac{1}{2}$ and the unit circle, we start with the equation of the unit circle, which is $x^2 + y^2 = 1$.

Substituting $y = \frac{1}{2}$ into the circle’s equation, we get:

$$x^2 + \left(\frac{1}{2}\right)^2 = 1$$

Simplifying this, we obtain:

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

We then isolate $x^2$:

$$x^2 = 1 – \frac{1}{4}$$

Simplifying further:

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

Taking the square root of both sides, we get:

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

Thus, the intersection points are:

$$(x, y) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$$ and $$(x, y) = \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)$$

Given an angle θ in the unit circle where sin(θ) = -1/2 and cos(θ) = -√3/2, find the coordinates (x, y) of the corresponding point on the unit circle Additionally, find the other angle φ in the range [0, 2π] which satisfies the same conditions

First, recognize that $\sin(\theta) = -\frac{1}{2}$ means that $\theta$ corresponds to either $210^\circ$ or $330^\circ$ in degrees, or $\frac{7\pi}{6}$ or $\frac{11\pi}{6}$ in radians. Since $\cos(\theta) = -\frac{\sqrt{3}}{2}$, we identify that $\theta$ must be in the third quadrant.

Therefore, the coordinates are:

$$ (x, y) = \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$

The angle $\phi$ in the range $[0, 2\pi]$ that satisfies the same conditions is $\frac{7\pi}{6}$.

Determine the angle in radians and the coordinates on the unit circle for the angle whose cosine is -1/2 in the interval [π, 2π], and find the corresponding sine value and tangent value

First, consider the unit circle where the cosine value is $\frac{-1}{2}$. In the interval $[\pi, 2\pi]$, the angle where the cosine is $\frac{-1}{2}$ is $\frac{4\pi}{3}$.

To find the corresponding sine value:

$$\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$

To find the tangent value:

$$\tan\left(\frac{4\pi}{3}\right) = \frac{\sin\left(\frac{4\pi}{3}\right)}{\cos\left(\frac{4\pi}{3}\right)} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$$

Therefore, the coordinates on the unit circle are:

$$\left(\cos\left(\frac{4\pi}{3}\right), \sin\left(\frac{4\pi}{3}\right)\right) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$$

Determine the Quadrant of a Point on the Unit Circle

$$\text{To determine the quadrant of a point } (x, y) \text{ on the unit circle, first consider the angle } \theta \text{ in radians.}$$

$$\text{For example, if } \theta = \frac{3\pi}{4}, \text{ we need to identify which quadrant this angle falls into.}$$

$$\text{Since } \theta \text{ is between } \frac{\pi}{2} \text{ and } \pi, \text{ it falls into the second quadrant. Therefore, the point is in Quadrant II.}$$

Find the coordinates of a point on the unit circle where the angle in standard position is given by 7π/6 radians

To solve this, recall that an angle of $\frac{7\pi}{6}$ radians is in the third quadrant.

The reference angle for $\frac{7\pi}{6}$ is $\frac{7\pi}{6} – \pi = \frac{\pi}{6}$.

The coordinates for $\frac{\pi}{6}$ on the unit circle are $\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$.

Because $\frac{7\pi}{6}$ is in the third quadrant, both coordinates are negative.

Thus, the coordinates are $$\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$$.