Math

Find the coordinates of the point on the unit circle at angle θ = π/4

The coordinates of the point on the unit circle at angle $ \theta = \frac{\pi}{4} $ can be found using the sine and cosine functions:

The x-coordinate is:

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

The y-coordinate is:

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

Therefore, the coordinates are:

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

How to remember the unit circle using trigonometric identities

To remember the unit circle, you can leverage trigonometric identities and properties:

1. Know the key angles and their corresponding coordinates: $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$, etc.

2. Understand that for any angle $\theta$, the coordinates on the unit circle are $(\cos\theta, \sin\theta)$.

3. Remember the symmetry properties: $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$.

4. Utilize special triangles (like $30^\circ-60^\circ-90^\circ$ and $45^\circ-45^\circ-90^\circ$) to derive coordinates.

With these strategies, you can reconstruct the unit circle efficiently.

Find the sine and cosine of 45 degrees in radians

To find the sine and cosine of $ \frac{\pi}{4} $, we use the unit circle. Since $ \frac{\pi}{4} $ corresponds to 45 degrees:

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

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

Prove the identity of sin(θ) on the unit circle

To prove the identity of $ \sin(\theta) $ on the unit circle, we start by considering a point on the unit circle at angle $ \theta $. The coordinates of this point can be represented as $ (\cos(\theta), \sin(\theta)) $.

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Using the Pythagorean identity for the unit circle, we have:

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

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Now consider a right triangle with the hypotenuse being the radius of the unit circle (which is 1). The opposite side of angle $ \theta $ is $ \sin(\theta) $ and the adjacent side is $ \cos(\theta) $.

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By the definition of sine in a right triangle, we get:

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$$ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $$

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Since the hypotenuse is 1, the opposite side is $ \sin(\theta) $, thus:

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

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This completes the proof.

How to calculate points on the unit circle for specific angles

To calculate points on the unit circle for a specific angle $ \theta $, follow these steps:

1. Recall that the unit circle is a circle with radius 1 centered at the origin (0,0).

2. Points on the unit circle are given by the coordinates $( \cos(\theta), \sin(\theta) )$, where $ \theta $ is the angle in radians measured from the positive x-axis.

3. For example, for $ \theta = \frac{ \pi }{4} $, the coordinates are:

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

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 $1$.