Math

Find the angle in radians and degrees for the point (-1/2, -√3/2) on the unit circle

We need to find the angle corresponding to the point $ \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) $ on the unit circle. This point lies in the third quadrant where both sine and cosine are negative. The reference angle is given by:

$$ \text{Reference angle} = \arccos\left( \frac{1}{2} \right) = \frac{\pi}{3} $$

Since the point is in the third quadrant, the angle in radians is:

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

To convert this to degrees:

$$ \theta = \frac{4\pi}{3} \times \frac{180}{\pi} = 240^{\circ} $$

Hence, the angle is $ \frac{4\pi}{3} $ radians or $ 240^{\circ} $.

Find the value of arcsin(1/2) in radians using the unit circle

To find the value of $ \arcsin(\frac{1}{2}) $, consider the unit circle and the definition of arcsin. The arcsin function outputs the angle whose sine is the given value within the range $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $.

For $ \arcsin(\frac{1}{2}) $, we need to find the angle $ \theta $ such that $ \sin(\theta) = \frac{1}{2} $. On the unit circle, $ \sin(30^{\circ}) = \frac{1}{2} $ or equivalently, in radians:

$$ \theta = \frac{\pi}{6} $$

Thus, the value of $ \arcsin(\frac{1}{2}) $ is:

$$ \arcsin(\frac{1}{2}) = \frac{\pi}{6} $$

Find the value of cosine at $\theta = \frac{3\pi}{4}$

The unit circle helps us locate the angle $\theta = \frac{3\pi}{4}$ which lies in the second quadrant. The reference angle for $\theta = \frac{3\pi}{4}$ is:

$$\pi – \frac{3\pi}{4} = \frac{\pi}{4}$$

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

$$\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right)$$

Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we have:

$$\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

Find the coordinates of the point where the angle π/4 intersects the unit circle

To find the coordinates of the point where the angle $ \frac{\pi}{4} $ intersects the unit circle, we use the unit circle definition. The unit circle has a radius of 1, and the coordinates of the points on the circle are given by the cosine and sine of the angle.

For the angle $ \frac{\pi}{4} $, we have:

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

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

Therefore, the coordinates are:

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

How to find the reference angle for any angle not on the unit circle

To find the reference angle for an angle θ not on the unit circle, you must first locate the angle in the appropriate quadrant. The reference angle is then the smallest angle between the terminal side of θ and the x-axis. Here are the steps:

1. If θ is in the first quadrant, the reference angle is θ itself:

$$ θ_{ref} = θ $$

2. If θ is in the second quadrant, the reference angle is:

$$ θ_{ref} = 180° – θ $$

3. If θ is in the third quadrant, the reference angle is:

$$ θ_{ref} = θ – 180° $$

4. If θ is in the fourth quadrant, the reference angle is:

$$ θ_{ref} = 360° – θ $$

Given that cos(θ) = -1/2, find the general solutions for θ in the unit circle

To solve for $ θ $ such that $ \cos(θ) = -\frac{1}{2} $, we need to find all angles in the unit circle where the cosine value is $ -\frac{1}{2} $. The cosine function is negative in the second and third quadrants.

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The reference angle for $ \cos(θ) = \frac{1}{2} $ is $ \frac{\pi}{3} $. Therefore, the general solutions in the second and third quadrants are:

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$$ θ = \pi – \frac{\pi}{3} + 2k\pi = \frac{2\pi}{3} + 2k\pi $$

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and

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$$ θ = \pi + \frac{\pi}{3} + 2k\pi = \frac{4\pi}{3} + 2k\pi $$

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where $ k $ is any integer.

How to memorize the coordinates of the unit circle

One method to memorize the unit circle is to remember key angles in radians and their corresponding coordinates. For example,:

$$ \text{At } \theta = 0, \ (1, 0) $$

$$ \text{At } \theta = \frac{\pi}{2}, \ (0, 1) $$

$$ \text{At } \theta = \pi, \ (-1, 0) $$

$$ \text{At } \theta = \frac{3\pi}{2}, \ (0, -1) $$

$$ \text{At } \theta = 2\pi, \ (1, 0) $$

These points divide the unit circle into four quadrants.

Prove that the integral of sin(x)/x from 0 to infinity is pi/2

To prove that the integral of $ \frac{\sin(x)}{x} $ from $ 0 $ to $ \infty $ is $ \frac{\pi}{2} $, we will use the fact that:

$$ \int_0^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} $$

The proof involves showing that the integral converges and evaluating it:

First, consider the function:

$$ f(t) = \int_0^t \frac{\sin(x)}{x} \, dx $$

As $ t \to \infty $, we must show that $ f(t) $ approaches $ \frac{\pi}{2} $. To do this, use the substitution $ x = t u $:

$$ \int_0^t \frac{\sin(x)}{x} \, dx = \int_0^1 \frac{\sin(tu)}{tu} \, t du = \int_0^1 \frac{\sin(tu)}{u} \, du $$

By integrating by parts and using properties of sine, we can show that:

$$ \int_0^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} $$

Solve for all x given that sin(x) + cos(2x) = 1, where x is an angle on the unit circle

To solve for all $x$ given that $\sin(x) + \cos(2x) = 1$:

First, use the double-angle identity for cosine: $\cos(2x) = 2\cos^2(x) – 1$

Substitute this into the equation:

$$ \sin(x) + 2\cos^2(x) – 1 = 1 $$

Rearrange the equation:

$$ \sin(x) + 2\cos^2(x) = 2 $$

Since $\sin^2(x) + \cos^2(x) = 1$, we have $\cos^2(x) = 1 – \sin^2(x)$

So:

$$ \sin(x) + 2(1 – \sin^2(x)) = 2 $$

Which simplifies to:

$$ \sin(x) + 2 – 2\sin^2(x) = 2 $$

Thus:

$$ \sin(x) – 2\sin^2(x) = 0 $$

Factor out $\sin(x)$:

$$ \sin(x)(1 – 2\sin(x)) = 0 $$

So $\sin(x) = 0$ or $\sin(x) = \frac{1}{2}$

Therefore, the solutions are:

$$ x = n\pi $$

and

$$ x = \frac{\pi}{6} + 2n\pi $$

or

$$ x = \frac{5\pi}{6} + 2n\pi $$

where $n$ is any integer.