Unit Circle

Calculate the exact values of the trigonometric functions for an angle of 7π/6 radians on the unit circle

To find the trigonometric functions for the angle $ \frac{7\pi}{6} $, locate the angle on the unit circle.

First, convert $ \frac{7\pi}{6} $ to degrees: $$ \frac{7\pi}{6} \times \frac{180^\circ}{\pi} = 210^\circ $$

Next, find the reference angle: $$ 210^\circ – 180^\circ = 30^\circ $$

Using the reference angle and the unit circle values, we have:

$$ \sin\left(\frac{7\pi}{6}\right) = -\sin\left(30^\circ\right) = -\frac{1}{2} $$

$$ \cos\left(\frac{7\pi}{6}\right) = -\cos\left(30^\circ\right) = -\frac{\sqrt{3}}{2} $$

$$ \tan\left(\frac{7\pi}{6}\right) = \frac{\sin\left(\frac{7\pi}{6}\right)}{\cos\left(\frac{7\pi}{6}\right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

Find the tangent value of π/4 in the unit circle

To find the tangent value of $ \frac{\pi}{4} $ in the unit circle, use the definition of tangent:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$

At $ \theta = \frac{\pi}{4} $, both the sine and cosine values are:

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

Therefore:

$$ \tan(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

Find the point(s) where the derivative of cos(theta) equals zero on the filled out unit circle

To find where the derivative of $ \cos(\theta) $ equals zero, we first need to find the derivative:

$$ \frac{d}{d\theta} \cos(\theta) = -\sin(\theta) $$

Set the derivative to zero:

$$ -\sin(\theta) = 0 $$

Thus, we have:

$$ \sin(\theta) = 0 $$

The solutions to this equation on the unit circle are:

$$ \theta = 0, \pi, 2\pi $$

Therefore, the points on the unit circle are:

$$ (1, 0), (-1, 0), (1, 0) $$

Find the equation for a unit circle in the Cartesian plane

The equation for a unit circle centered at the origin in the Cartesian plane is:

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

This equation represents all points $(x, y)$ that are exactly one unit away from the origin.

Find the value of tan(π/4) using the unit circle

To find the value of $ \tan(\frac{\pi}{4}) $ using the unit circle:

On the unit circle, the coordinates for $ \frac{\pi}{4} $ are $ (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $.

Therefore:

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

Find the value of sec(θ) for θ = π/4 on the unit circle

To find the value of $ \sec(\theta) $ for $ \theta = \frac{\pi}{4} $ on the unit circle, we use the definition of secant, which is the reciprocal of cosine:

$$ \sec(\theta) = \frac{1}{\cos(\theta)} $$

For $ \theta = \frac{\pi}{4} $, we have:

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

Therefore:

$$ \sec\left(\frac{\pi}{4}\right) = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} $$

Find the point on the unit circle where the sine value is negative and the cosine value is positive

The unit circle is defined as the set of all points $(x, y)$ such that:

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

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In the unit circle, the sine value corresponds to the y-coordinate and the cosine value corresponds to the x-coordinate. We need to find a point where:

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$$ y < 0 $$

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$$ x > 0 $$

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One such point is:

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

Determine the coordinates of points on the unit circle that satisfy the given equation

To determine the coordinates of points on the unit circle that satisfy the equation $ \cos^2(\theta) – \sin^2(\theta) = 0 $:

First, we recall the Pythagorean identity: $$ \cos^2(\theta) + \sin^2(\theta) = 1 $$

Given the equation: $$ \cos^2(\theta) – \sin^2(\theta) = 0 $$

This can be rewritten as: $$ \cos^2(\theta) = \sin^2(\theta) $$

Taking the square root of both sides gives: $$ \cos(\theta) = \pm \sin(\theta) $$

We consider the positive and negative cases separately.

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

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

Where $ k $ is any integer.

For $ \cos(\theta) = -\sin(\theta) $:

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

Where $ k $ is any integer.

Thus, the coordinates of the points on the unit circle are:

$$ ( \cos(\frac{\pi}{4} + k \pi), \sin(\frac{\pi}{4} + k \pi) ) $$

$$ ( \cos(\frac{3\pi}{4} + k \pi), \sin(\frac{3\pi}{4} + k \pi) ) $$

What is the value of cos(π/4) on the unit circle?

Find the coordinates of the point on the unit circle where the inverse sine function is equal to 1/2

To find the coordinates of the point on the unit circle where the inverse sine function, $ \sin^{-1}(x) $, is equal to $ \frac{1}{2} $:

We need to solve the equation:

$$ \sin^{-1}(y) = \frac{1}{2} $$

The angle whose sine is $ \frac{1}{2} $ is:

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

On the unit circle, the coordinates corresponding to this angle are:

$$ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) $$

Evaluating the trigonometric functions, we get:

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

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

So, the coordinates are:

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