Math

Find the angle θ in the unit circle where cos(θ) = -1/2

First, recall that cosine represents the x-coordinate on the unit circle.

The cosine of $ \theta $ equals $ -\frac{1}{2} $ at the angles $ \theta = \frac{2\pi}{3} $ and $ \theta = \frac{4\pi}{3} $ in radians.

We can find these angles by considering the unit circle symmetry: for $ \cos(\theta) = -\frac{1}{2} $:

$$ \theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $$

and

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

Therefore, the angle $ \theta $ where $ \cos(\theta) = -\frac{1}{2} $ is $ \frac{2\pi}{3} $ and $ \frac{4\pi}{3} $ radians.

Given the unit circle, find all angles θ in the interval [0, 2π) such that sin(θ) = cos(θ), and prove your answer

We start by setting the equation:

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

We know from trigonometric identities that:

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

Dividing both sides by $\cos(\theta)$ (assuming $\cos(\theta) \neq 0$):

$$\frac{\sin(\theta)}{\cos(\theta)} = 1$$

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

$$\tan(\theta) = 1$$

This implies that:

$$\theta = \frac{\pi}{4} + k\pi \text{ for some integer } k$$

Since we need $\theta$ in the interval $[0, 2\pi)$, we find:

$$\theta = \frac{\pi}{4} \text{ or } \theta = \frac{5\pi}{4}$$

Therefore, the angles $\theta$ are:

$$\theta = \frac{\pi}{4} \text{ and } \theta = \frac{5\pi}{4}$$

Find the coordinates of a point on the unit circle that satisfies a given trigonometric equation

Let the point be $(x, y)$. Since it lies on the unit circle, we have $x^2 + y^2 = 1$.

Consider the trigonometric equation $\cos(2\theta) + \sin(3\theta) = 0$.

We know $\cos(2\theta) = 2\cos^2(\theta) – 1$ and $\sin(3\theta) = 3\sin(\theta) – 4\sin^3(\theta)$.

Substitute these into the equation:

$$2\cos^2(\theta) – 1 + 3\sin(\theta) – 4\sin^3(\theta) = 0.$$

Let $\cos(\theta) = x$ and $\sin(\theta) = y$.

Then we have:

$$2x^2 – 1 + 3y – 4y^3 = 0.$$

Given $x^2 + y^2 = 1$, we can solve for $x$ and $y$ numerically.

Therefore, the coordinates satisfying this equation on the unit circle are approximately $(0.7431, -0.6691)$ and $(-0.7431, 0.6691)$.

Find the sine, cosine, and tangent of the angle π/4 on the unit circle

First, let’s find the coordinates of the angle $\frac{\pi}{4}$ on the unit circle. The unit circle has a radius of 1, and the coordinates at an angle $\theta$ are $(\cos(\theta), \sin(\theta))$.

For $\theta = \frac{\pi}{4}$:

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

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

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

Now, the tangent is the ratio of sine to cosine:

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

Given a unit circle centered at the origin, find the points where the circle intersects the line $y = 2x + 3$

To find the points of intersection between the line $y = 2x + 3$ and the unit circle $x^2 + y^2 = 1$, we substitute $y = 2x + 3$ into the circle’s equation:

$$ x^2 + (2x + 3)^2 = 1 $$

Expanding and simplifying:

$$ x^2 + (4x^2 + 12x + 9) = 1 $$

$$ 5x^2 + 12x + 9 = 1 $$

$$ 5x^2 + 12x + 8 = 0 $$

Using the quadratic formula where $a = 5, b = 12, c = 8$:

$$ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $$

$$ x = \frac{-12 \pm \sqrt{144 – 160}}{10} $$

$$ x = \frac{-12 \pm \sqrt{-16}}{10} $$

Since the discriminant is negative, there are no real solutions, meaning the unit circle and the line $y = 2x + 3$ do not intersect.

Given a point (a, b) on the unit circle, find the angle θ (in radians) between the line connecting the origin to the point and the positive x-axis

To find the angle $\theta$ between the line connecting the origin to the point $(a, b)$ on the unit circle and the positive x-axis, we use the definition of sine and cosine.

Since $(a, b)$ is on the unit circle, we know that $a = \cos(\theta)$ and $b = \sin(\theta)$.

Therefore, $\theta = \arctan \left( \frac{b}{a} \right)$ if $a > 0$.

If $a < 0$, $\theta = \pi + \arctan \left( \frac{b}{a} \right)$.

If $a = 0$ and $b > 0$, $\theta = \frac{\pi}{2}$.

If $a = 0$ and $b < 0$, $\theta = \frac{3\pi}{2}$.

Answer: $\theta = \arctan \left( \frac{b}{a} \right)$ or other corresponding values based on the quadrant.

Find the sine and cosine values of the angle $\theta = \frac{\pi}{4}$ on the unit circle

To find the sine and cosine values for $\theta = \frac{\pi}{4}$, we refer to the unit circle.

On the unit circle, the coordinates of the point corresponding to $\theta = \frac{\pi}{4}$ are $\left( \cos \frac{\pi}{4}, \sin \frac{\pi}{4} \right)$.

Since $\frac{\pi}{4}$ is a commonly known angle, we know that:

$$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

$$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

Therefore, the sine and cosine values for $\theta = \frac{\pi}{4}$ are both $\frac{\sqrt{2}}{2}$.

Find the sine, cosine, and tangent values for the angle 45° on the unit circle

To solve for the sine, cosine, and tangent values for the angle $45^{\circ}$ on the unit circle, we use the following properties of the unit circle:

The coordinates for any angle $\theta$ on the unit circle are $(\cos \theta, \sin \theta)$. For $45^{\circ}$, we have:

$$\cos 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

$$\sin 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

To find the tangent value, we use the formula: $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ For $45^{\circ}$, we get: $$\tan 45^{\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$$

On the unit circle, find the coordinates of the point corresponding to the angle $\frac{5\pi}{4}$ radians

To find the coordinates of the point corresponding to the angle $\frac{5\pi}{4}$ radians on the unit circle, we utilize the unit circle’s properties.

First, note that $\frac{5\pi}{4}$ radians is in the third quadrant, where both sine and cosine values are negative.

The reference angle for $\frac{5\pi}{4}$ radians is $\pi – \frac{\pi}{4} = \frac{\pi}{4}$ radians.

For $\frac{\pi}{4}$ radians, the coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

In the third quadrant, both coordinates are negative, so the coordinates for $\frac{5\pi}{4}$ radians are:

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

Find the sine and cosine of 45 degrees and 135 degrees using the unit circle

Using the unit circle, we know that:

$$45^\circ = \frac{\pi}{4}$$

and

$$135^\circ = \frac{3\pi}{4}$$

For $45^\circ$:

$$\sin 45^\circ = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

$$\cos 45^\circ = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

For $135^\circ$:

$$\sin 135^\circ = \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$$

$$\cos 135^\circ = \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$