Unit Circle

Find the value of cos(π/4) and sin(π/4) using the unit circle

To find the values of $ \cos(\frac{\pi}{4}) $ and $ \sin(\frac{\pi}{4}) $ using the unit circle, we need to identify the coordinate point on the unit circle that corresponds to the angle $ \frac{\pi}{4} $.

The angle $ \frac{\pi}{4} $ is located in the first quadrant where both sine and cosine values are positive. This angle corresponds to the point $ (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $ on the unit circle.

Therefore:

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

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

Find the equation of the tangent line to the unit circle at (1, 0)

The unit circle is given by the equation:

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

To find the equation of the tangent line at $(1, 0)$, we first find the slope of the tangent. Differentiate the equation implicitly with respect to $x$:

$$ 2x + 2y \x0crac{dy}{dx} = 0 $$

At the point $(1, 0)$, substitute $x = 1$ and $y = 0$:

$$ 2(1) + 2(0) \x0crac{dy}{dx} = 0 $$

So, the slope $\x0crac{dy}{dx}$ at $(1, 0)$ is $0$. The equation of the tangent line using point-slope form is:

$$ y – 0 = 0(x – 1) $$

Therefore, the equation is:

$$ y = 0 $$

Find the value of tan(θ) using the unit circle when θ is in the third quadrant

To find the value of $ \tan(θ) $ using the unit circle, we need to determine the coordinates where $ θ $ intersects the unit circle in the third quadrant.

In the third quadrant, both the x and y coordinates are negative. Suppose $ θ = 225° $ (or $ \frac{5π}{4} $ in radians). In this case, the coordinates on the unit circle are $ ( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} ) $.

The tangent of $ θ $ is given by the ratio of the y-coordinate to the x-coordinate:

$$ \tan(225°) = \frac{y}{x} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 $$

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:

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

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

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 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

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

Find the sine and cosine of the angle π/3 using the unit circle

To find the sine and cosine of the angle $ \pi/3 $ using the unit circle, consider the angle that corresponds to $ \pi/3 $ radians (or 60 degrees).

In the unit circle, the coordinates of the point on the circumference corresponding to the angle $ \pi/3 $ are $ (\cos(\pi/3), \sin(\pi/3)) $.

For $ \pi/3 $:

$$ \cos(\pi/3) = \frac{1}{2} $$

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

Find the coordinates of the point where the terminal side of the angle intersects the unit circle at an angle of 5π/4 radians

To find the coordinates of the point where the terminal side of the angle intersects the unit circle at an angle of $\frac{5\pi}{4}$ radians, we use the unit circle properties.

The angle $\frac{5\pi}{4}$ radians is in the third quadrant where both sine and cosine values are negative.

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

The coordinates on the unit circle for an angle of $\frac{\pi}{4}$ are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

Since we are in the third quadrant, we change the signs of both x and y coordinates:

Therefore, the coordinates are:

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