Unit Circle

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

Consider the angle $ \theta = \frac{3\pi}{4} $, which is in the second quadrant.

Using the unit circle, we know:

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

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

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

Find the exact trigonometric values of cos(5π/6) and sin(5π/6) from the unit circle

To find the exact values of $\cos\left(\frac{5\pi}{6}\right)$ and $\sin\left(\frac{5\pi}{6}\right)$, we refer to the unit circle.

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

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

On the unit circle, the coordinates for $\frac{\pi}{6}$ are $(\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6}))$ = (\frac{\sqrt{3}}{2}, \frac{1}{2})$

Since $\frac{5\pi}{6}$ is in the second quadrant, $\cos(\frac{5\pi}{6})$ is negative and $\sin(\frac{5\pi}{6})$ is positive:

Thus, $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$ and $\sin(\frac{5\pi}{6}) = \frac{1}{2}$

Find the coordinates of a point on the unit circle at an angle of π/6

To find the coordinates of a point on the unit circle at an angle of $ \frac{\pi}{6} $, we use the fact that the coordinates are given by $ ( \cos(\theta), \sin(\theta)) $ where $ \theta $ is the angle:

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

Therefore:

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

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

The coordinates are:

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

Calculate the area of the shaded region in a unit circle with central angles

Let’s calculate the area of the shaded region in a unit circle with central angles $ \theta $ and $ \alpha $.

The area of a sector of a circle is given by:

$$ A = \frac{1}{2} r^2 \theta $$

For a unit circle, $ r = 1 $, so the above formula simplifies to:

$$ A = \frac{1}{2} \theta $$

The area of the shaded region is then the difference between two sector areas:

$$ A_{shaded} = \frac{1}{2} (\theta – \alpha) $$

Find the coordinates on the unit circle for the angles

Find the coordinates on the unit circle for the angle $ \theta = \frac{\pi}{4} $:

The coordinates are given by $ (\cos(\theta), \sin(\theta)) $.

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

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

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

The coordinates are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

Find the tangent values at 0, π/4, and π/3 on the unit circle

To find the tangent values at points $0$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$ on the unit circle, we use the tangent function $tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$:

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 $\theta = \frac{\pi}{3}$:

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

Find the coordinates where $ \cos(\theta) = \sin(\theta) $ on the unit circle

To find the coordinates where $ \cos(\theta) = \sin(\theta) $ on the unit circle, we start from the equation:

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

Since both cosine and sine are equal, we can express this as:

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

Divide both sides by $ \cos(\theta) $:

$$1 = \tan(\theta) $$

This implies

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

for integer values of n. The corresponding coordinates on the unit circle are:

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

Calculate the exact value of sin(7π/6) using the unit circle

To determine the exact value of $\sin\left(\frac{7\pi}{6}\right)$ using the unit circle, first note that $\frac{7\pi}{6}$ is in the third quadrant.

In the third quadrant, the sine function is negative.

Now, find the reference angle for $\frac{7\pi}{6}$:

$$ 7\pi / 6 – \pi = \pi / 6 $$

The reference angle is $\pi / 6$, whose sine value is $\frac{1}{2}$.

Since sine is negative in the third quadrant:

$$ \sin \left( \frac{7\pi}{6} \right) = -\frac{1}{2} $$

Find the value of arcsin(sqrt(3)/2) based on the unit circle

To find the value of $ \arcsin(\sqrt{3}/2) $, we need to locate where $ \sin(x) = \sqrt{3}/2 $ on the unit circle.

On the unit circle, $ \sin(x) = \sqrt{3}/2 $ at the angles:

$$ x = \frac{\pi}{3} $$

and

$$ x = \frac{2\pi}{3} $$

So,

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

in the primary range of arcsin, which is $ [ -\frac{\pi}{2}, \frac{\pi}{2} ] $.

Find the value of arccos(-1/2) and verify it using the unit circle

To find the value of $\arccos(-1/2)$, we need to determine the angle in the unit circle whose cosine is $-1/2$.

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From the unit circle, we know that:

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

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Hence, the value of $\arccos(-1/2)$ is $\frac{2\pi}{3}$.

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Verification:

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Consider the angle $\frac{2\pi}{3}$ in the unit circle, its cosine value is:

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

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This matches our original value, verifying that $\arccos(-1/2) = \frac{2\pi}{3}$.