Unit Circle

Given a point (x,y) on the unit circle where the angle in radians is \(\theta\), express the coordinates of the point in terms of \(\cos(\theta)\) and \(\sin(\theta)\), and show how these expressions can be derived from the Pythagorean identity Additiona

The unit circle is defined by the equation:

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

Since the point \((x,y)\) lies on the unit circle, we can express \(x\) and \(y\) in terms of \(\cos(\theta)\) and \(\sin(\theta)\):

$$ x = \cos(\theta) $$

$$ y = \sin(\theta) $$

These expressions satisfy the Pythagorean identity:

$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$

Now, given \( \cos(\theta) = -\frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \), we need to find the angle \(\theta\).

The values correspond to the angle \( \theta = \frac{2\pi}{3} \) or \( \theta = \frac{4\pi}{3} \) (in the second and third quadrants respectively where cosine is negative and sine is positive).

Therefore, the angles in radians are:

$$ \theta = \frac{2\pi}{3}, \frac{4\pi}{3} $$

Find the value of sin, cos, and tan for 45 degrees using the unit circle

To find the values of $\sin$, $\cos$, and $\tan$ for $45^\circ$ using the unit circle, we first note that $45^\circ$ is equivalent to $\frac{\pi}{4}$ radians.

On the unit circle, the coordinates for $\frac{\pi}{4}$ radians (or $45^\circ$) are $( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} )$.

Therefore:

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

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

$$\tan(45^\circ) = \frac{\sin(45^\circ)}{\cos(45^\circ)} = 1 $$

Consider a unit circle with center at the origin A square is inscribed in the circle, and a triangle is inscribed in the square Determine the area of the triangle

$$\text{First, determine the side length of the square inscribed in the unit circle. The diagonal of the square is equal to the diameter of the circle, which is 2.}$$

$$\text{Using Pythagoras’ theorem, the side length } a \text{ of the square is given by:}$$

$$a\sqrt{2} = 2$$

$$a = \frac{2}{\sqrt{2}} = \sqrt{2}$$

$$\text{Next, consider an equilateral triangle inscribed in the square. The side length of the triangle is the same as the side length of the square, } a = \sqrt{2}.\text{ The area of an equilateral triangle with side length } a \text{ is given by: }$$

$$A = \frac{\sqrt{3}}{4} a^2$$

$$A = \frac{\sqrt{3}}{4} (\sqrt{2})^2 = \frac{\sqrt{3}}{4} \times 2 = \frac{\sqrt{3}}{2}$$

$$\text{Therefore, the area of the triangle is } \frac{\sqrt{3}}{2}. $$

Find the coordinates on the unit circle at different angles

To find the coordinates on the unit circle at 45 degrees:

The unit circle equation is given by:

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

For an angle of \(45^{\circ}\) or \(\frac{\pi}{4}\) radians, the coordinates are:

$$x = \cos(\frac{\pi}{4})$$

$$y = \sin(\frac{\pi}{4})$$

Both \(\cos(\frac{\pi}{4})\) and \(\sin(\frac{\pi}{4})\) are equal to \(\frac{\sqrt{2}}{2}\).

Hence, the coordinates are:

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

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.