Math

Determine the exact values of sine and cosine for an angle of 5π/6 in the unit circle

To determine the exact values of $\sin(\frac{5\pi}{6})$ and $\cos(\frac{5\pi}{6})$, we use the unit circle.

For the angle $\frac{5\pi}{6}$, it is in the second quadrant where sine is positive and cosine is negative. The reference angle for $\frac{5\pi}{6}$ is:

$$ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $$

The values for sine and cosine at $\frac{\pi}{6}$ are known:

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

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

Since $\frac{5\pi}{6}$ is in the second quadrant:

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

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

Given a point on the unit circle, find the corresponding angle and verify the coordinates

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Determine the value of tan(θ) at θ = 3π/4 using the unit circle chart

To determine the value of $ \tan(\theta) $ at $ \theta = \frac{3\pi}{4} $, we use the unit circle chart. The angle $ \frac{3\pi}{4} $ is in the second quadrant, where the reference angle is $ \frac{\pi}{4} $. In this quadrant, the tangent value is negative.

Since $ \tan(\frac{\pi}{4}) = 1 $,

$$ \tan(\frac{3\pi}{4}) = -1 $$

Therefore, the value of $ \tan(\frac{3\pi}{4}) $ is:

$$ \boxed{-1} $$

Find the coordinates of a point on the unit circle at angle pi/3

The unit circle has a radius of 1. The coordinates of a point on the circle at angle $ \frac{\pi}{3} $ are given by:

$$ (\cos(\frac{\pi}{3}), \sin(\frac{\pi}{3})) $$

Therefore,

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

and

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

Thus, the coordinates are:

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

Determine the values of sin(θ), cos(θ), and tan(θ) for θ in the second quadrant of the unit circle

In the second quadrant, the angle $ \theta $ ranges from $ \frac{\pi}{2} $ to $ \pi $. Here, $ \sin(\theta) $ is positive, $ \cos(\theta) $ is negative, and $ \tan(\theta) $ is negative.

Using the unit circle, for $ \theta = \frac{2\pi}{3} $:

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

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

$$ \tan(\frac{2\pi}{3}) = -\sqrt{3} $$

Find the value of sin(π/3) on the unit circle

To find the value of $ \sin\left(\frac{\pi}{3}\right) $ on the unit circle, we look at the reference angle for $ \frac{\pi}{3} $.

The reference angle is $ 60^{\circ} $, and the sine value for $ 60^{\circ} $ is:

$$ \sin\left(60^{\circ}\right) = \frac{\sqrt{3}}{2} $$

Therefore:

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

Find the area of a sector of a unit circle with a central angle of θ radians

To find the area of a sector of a unit circle with a central angle of $ \theta $, we use the formula for the area of a sector:

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

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Since the radius $ r $ of a unit circle is 1, the formula simplifies to:

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

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The area of the sector is:

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

Find the value of sin(π/6), cos(π/3), and tan(π/4) using the unit circle chart

Using the unit circle chart, we find:

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

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

$$ \tan\left(\frac{\pi}{4}\right) = 1 $$

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:

$$ y = 0 $$