Unit Circle

Given that tan(θ) = 2 and θ is in the second quadrant, find the exact values of sin(θ) and cos(θ)

1. Given that $ \tan(\theta) = 2 $, we can write:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = 2 $$

Let $ \sin(\theta) = 2k $ and $ \cos(\theta) = -k $ (since $ \theta $ is in the second quadrant where cosine is negative). Then:

$$ \frac{2k}{-k} = 2 $$

2. From the Pythagorean identity:

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

Substitute the values:

$$ (2k)^2 + (-k)^2 = 1 $$

$$ 4k^2 + k^2 = 1 $$

3. Solving for $ k $:

$$ 5k^2 = 1 $$

$$ k^2 = \frac{1}{5} $$

$$ k = \pm \frac{1}{\sqrt{5}} $$

4. Since $ \sin(\theta) = 2k $ and $ \cos(\theta) = -k $, we have:

$$ \sin(\theta) = 2 \left( \frac{1}{\sqrt{5}} \right) = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} $$

$$ \cos(\theta) = – \left( \frac{1}{\sqrt{5}} \right) = -\frac{\sqrt{5}}{5} $$

Find the cosine of 2π/3 radians on the unit circle

The angle $ \frac{2\pi}{3} $ radians is in the second quadrant.

In the second quadrant, the cosine of an angle is negative.

For $ \frac{2\pi}{3} $ radians, the reference angle is $ \frac{\pi}{3} $ radians.

Cosine of $ \frac{\pi}{3} $ radians is $ \frac{1}{2} $.

Therefore, the cosine of $ \frac{2\pi}{3} $ radians is:

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

Prove that the integral of exp(i*theta) over a complete unit circle is zero

To prove that the integral of $ \exp(i \theta) $ over a complete unit circle is zero, we evaluate the contour integral:

$$ \int_0^{2\pi} \exp(i \theta) d\theta $$

Recall that $ \exp(i \theta) = \cos(\theta) + i \sin(\theta) $. So the integral becomes:

$$ \int_0^{2\pi} \cos(\theta) d\theta + i \int_0^{2\pi} \sin(\theta) d\theta $$

We know that the integrals of $ \cos(\theta) $ and $ \sin(\theta) $ over a complete period from $ 0 $ to $ 2 \pi $ are both zero:

$$ \int_0^{2\pi} \cos(\theta) d\theta = 0 $$

$$ \int_0^{2\pi} \sin(\theta) d\theta = 0 $$

Thus, the original integral evaluates to:

$$ \int_0^{2\pi} \exp(i \theta) d\theta = 0 $$

Find the sine and cosine values for an angle of π/3 on the unit circle

To find the sine and cosine values for an angle of $ \frac{π}{3} $ on the unit circle, we need to recall the special angles on the unit circle.

For $ \frac{π}{3} $, the coordinates are (cos($ \frac{π}{3} $), sin($ \frac{π}{3} $)).

Using the unit circle,

$$ \cos(\frac{π}{3}) = \frac{1}{2} $$

$$ \sin(\frac{π}{3}) = \frac{\sqrt{3}}{2} $$

So, the sine and cosine values for $ \frac{π}{3} $ are:

$$ \cos(\frac{π}{3}) = \frac{1}{2} $$

$$ \sin(\frac{π}{3}) = \frac{\sqrt{3}}{2} $$

Do unit circles have diameter of 1

A unit circle is defined as a circle with a radius of $1$ unit.

The diameter of a circle is twice the radius.

Therefore, for a unit circle:

$$ \text{Diameter} = 2 \times \text{Radius} = 2 \times 1 = 2 $$

Thus, the diameter of a unit circle is $2$ units, not $1$.

Find the tangent values at 0, π/4, and π/2 on the unit circle

To find the tangent values at specific points on the unit circle, we use the definition of the tangent function, which is $$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$.

1. At $ \theta = 0 $:

$$ \tan(0) = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0 $$

2. At $ \theta = \frac{\pi}{4} $:

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

3. At $ \theta = \frac{\pi}{2} $:

$$ \tan\left(\frac{\pi}{2}\right) = \frac{\sin\left(\frac{\pi}{2}\right)}{\cos\left(\frac{\pi}{2}\right)} = \frac{1}{0} $$

Since division by zero is undefined, $ \tan\left(\frac{\pi}{2}\right) $ does not exist.

Find the coordinates of the point on the unit circle where the angle is pi over 4 radians

To find the coordinates of the point on the unit circle where the angle is $ \frac{\pi}{4} $ radians, we use the unit circle definition:

The coordinates at this angle are:

$$ \left( \cos \frac{\pi}{4} , \sin \frac{\pi}{4} \right) $$

Using known values:

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

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

So, the coordinates are:

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

Identify the coordinates on the unit circle for angle θ

The coordinates of a point on the unit circle for a given angle $ \theta $ are given by:

$$ ( \cos(\theta), \sin(\theta) ) $$

For example, if $ \theta = \frac{\pi}{4} $:

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

Find the values of tan(θ) for θ in the interval [0, 2π] that satisfy the equation tan(θ) = 2

To find the values of $ \tan(\theta) $ that satisfy the equation $ \tan(\theta) = 2 $ in the interval $ [0, 2\pi] $, we need to determine the angles where the tangent function equals 2.

First, recall that the tangent function is periodic with period $ \pi $, and the angles where $ \tan(\theta) = 2 $ are:

$$ \theta_1 = \arctan(2) $$

and

$$ \theta_2 = \arctan(2) + \pi $$

Because the tangent function repeats every $ \pi $ radians, we only need to check within one period:

$$ \theta_1 = \arctan(2) $$

$$ \theta_2 = \arctan(2) + \pi $$

Thus, the solutions within $ [0, 2\pi] $ are:

$$ \theta = \arctan(2) $$

and

$$ \theta = \arctan(2) + \pi $$

Solve the equation to find all distinct points where the ellipse intersects the unit circle

To find where the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intersects the unit circle $x^2 + y^2 = 1$, we need to solve the system of equations:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

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

First, substitute $y^2 = 1 – x^2$ into the ellipse equation:

$$ \frac{x^2}{a^2} + \frac{1 – x^2}{b^2} = 1 $$

Multiply through by $a^2b^2$ to clear the denominators:

$$ b^2x^2 + a^2(1 – x^2) = a^2b^2 $$

Simplify and solve for $x^2$:

$$ b^2x^2 + a^2 – a^2x^2 = a^2b^2 $$

$$ (b^2 – a^2)x^2 = a^2(b^2 – 1) $$

$$ x^2 = \frac{a^2(b^2 – 1)}{b^2 – a^2} $$

Then solve for $y^2$ using $y^2 = 1 – x^2$.