Math

Find the point of intersection of a line passing through the origin at an angle θ with the unit circle

To find the point of intersection of a line passing through the origin at an angle $ \theta $ with the unit circle, we start by writing the equation of the line. The equation of the line in the form $ y = mx $ is:

$$ y = \tan(\theta) x $$

Since the line intersects the unit circle, we substitute $ y = \tan(\theta) x $ into the equation of the unit circle $ x^2 + y^2 = 1 $:

$$ x^2 + (\tan(\theta) x)^2 = 1 $$

Simplifying, we have:

$$ x^2 + x^2 \tan^2(\theta) = 1 $$

$$ x^2(1 + \tan^2(\theta)) = 1 $$

We use the trigonometric identity $ 1 + \tan^2(\theta) = \sec^2(\theta) $:

$$ x^2 \sec^2(\theta) = 1 $$

$$ x^2 = \cos^2(\theta) $$

So, we get two possible values for $ x $:

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

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

For each $ x $, we find the corresponding $ y $:

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

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

When $ x = -\cos(\theta) $:

$$ y = \tan(\theta) (-\cos(\theta)) = -\sin(\theta) $$

So, the points of intersection are:

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

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

Find the angle θ on the unit circle where the equation cos^2(θ) – sin^2(θ) = 1 – 2sin^2(θ) holds true

To solve for $ \theta $ on the unit circle in the equation $ \cos^2(\theta) – \sin^2(\theta) = 1 – 2\sin^2(\theta) $, start by using trigonometric identities:

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We know that $ \cos^2(\theta) = 1 – \sin^2(\theta) $, so the equation becomes:

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$$ 1 – \sin^2(\theta) – \sin^2(\theta) = 1 – 2\sin^2(\theta) $$

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Simplify both sides:

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$$ 1 – 2\sin^2(\theta) = 1 – 2\sin^2(\theta) $$

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The equation holds for any $ \theta $ where $ 1 – 2\sin^2(\theta) $ is defined, which simplifies to $ \theta = n\pi $, where $ n $ is an integer.

Find the exact values of sin(θ), cos(θ), and tan(θ) at θ = 3π/4

To find the exact values of $ \sin(\theta) $, $ \cos(\theta) $, and $ \tan(\theta) $ at $ \theta = \frac{3\pi}{4} $, we use the unit circle:

For $ \theta = \frac{3\pi}{4} $, the corresponding point on the unit circle is in the second quadrant where both $ \sin(\theta) $ and $ \cos(\theta) $ have specific values:

$ \sin(\frac{3\pi}{4}) $: The sine value is given by:

$$ \sin(\frac{3\pi}{4}) = \sin(\pi – \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$

$ \cos(\frac{3\pi}{4}) $: The cosine value is given by:

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

$ \tan(\frac{3\pi}{4}) $: The tangent value is given by:

$$ \tan(\frac{3\pi}{4}) = \frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

Find the exact coordinates of the point where the angle 7π/6 intersects the unit circle

To find the coordinates of the point where the angle $ \frac{7\pi}{6} $ intersects the unit circle, we first identify the reference angle. The reference angle for $ \frac{7\pi}{6} $ is $ \frac{\pi}{6} $.

The coordinates for the angle $ \frac{\pi}{6} $ on the unit circle are $ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $.

Since $ \frac{7\pi}{6} $ is in the third quadrant, both x and y coordinates will be negative:

$$ \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$

Therefore, the coordinates are $ \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $.

Find the coordinates of the point where the terminal side of theta intersects the unit circle at theta = 5π/6

To find the coordinates of the point where the terminal side of $ \theta $ intersects the unit circle at $ \theta = \frac{5\pi}{6} $, we use the unit circle definition and the corresponding reference angle.

The reference angle for $ \theta = \frac{5\pi}{6} $ is $ \frac{\pi}{6} $. The coordinates on the unit circle for $ \frac{\pi}{6} $ are $ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $.

Since $ \frac{5\pi}{6} $ is in the second quadrant, we adjust the signs of the coordinates:

$$ \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) $$

Find the sine value for 5π/6 on the unit circle

To find the sine value for $ \frac{5\pi}{6} $ on the unit circle, we follow these steps:

First, understand that $ \frac{5\pi}{6} $ is in the second quadrant.

The reference angle is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.

In the second quadrant, sine is positive, and we know:

$$ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} $$

Therefore,

$$ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} $$

Find the value of sine and cosine for specific angles using the unit circle

Using the unit circle, find the values of $\sin$ and $\cos$ for the angle $\frac{\pi}{4}$.

For $\theta = \frac{\pi}{4}$:

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

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

Identify the angle on the unit circle for cos(θ) = 1/2

To identify the angles where $\cos(\theta) = \frac{1}{2}$, we look at the unit circle:

In the first quadrant, $\theta = \frac{\pi}{3}$, and in the fourth quadrant, $\theta = \frac{5\pi}{3}$.

Determine the cosine value of an angle given on the unit circle

Given an angle $ \theta $ on the unit circle, we need to determine the value of $ \cos(\theta) $.

For example, if $ \theta = \frac{\pi}{3} $, we can use the unit circle to find:

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

Thus, $ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} $.

Find the coordinates of a point on the unit circle corresponding to an angle of 5π/6

To find the coordinates of a point on the unit circle at an angle of $ \frac{5\pi}{6} $, we use the unit circle definitions for sine and cosine:

$$ \text{cos}(\theta) = \text{x-coordinate} $$

$$ \text{sin}(\theta) = \text{y-coordinate} $$

For $ \frac{5\pi}{6} $:

$$ \text{cos}(\frac{5\pi}{6}) = – \frac{\sqrt{3}}{2} $$

$$ \text{sin}(\frac{5\pi}{6}) = \frac{1}{2} $$

So, the coordinates are:

$$ \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) $$