Math

Determine the tangent values for points on the unit circle at angles θ = π/4, θ = 2π/3, and θ = 5π/6

To find the tangent values for the given angles on the unit circle, we use the definition of the tangent function, which is $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $.

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

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

Therefore,

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

For $ \theta = \frac{2\pi}{3} $:

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

Therefore,

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

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

$$ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \text{ and } \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} $$

Therefore,

$$ \tan\left(\frac{5\pi}{6}\right) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$

Calculate the coordinates of a point on the unit circle, given theta in radians

To calculate the coordinates of a point on the unit circle given an angle $\theta$ in radians, use the formulas:

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

and

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

For example, if $\theta = \frac{\pi}{4}$, then:

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

and

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

Therefore, the coordinates are:

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

Find the coordinates where the function f(theta) = sin(2theta) intersects with the unit circle

To find the coordinates where $ f(\theta) = \sin(2\theta) $ intersects the unit circle, we start by setting $ \sin(2\theta) = y $.

The unit circle equation is $ x^2 + y^2 = 1 $.

Since $ y = \sin(2\theta) $, we have $ x^2 + \sin^2(2\theta) = 1 $.

Using the double angle identity, $ \sin(2\theta) = 2\sin(\theta)\cos(\theta) $, we rewrite the equation:

$$ x^2 + 4\sin^2(\theta)\cos^2(\theta) = 1 $$

Next, let $ u = \sin(\theta) $ and $ v = \cos(\theta) $ so the equation becomes:

$$ x^2 + 4uv = 1 $$

We need to satisfy both $ u^2 + v^2 = 1 $ and $ x^2 + 4uv = 1 $. Solving for $ x $ and substituting values:

After solving, we find the coordinates where $ f(\theta) $ intersects the unit circle are:

$$ (x_1, y_1) = (\sqrt{1 – \sin^2(2\theta)}, \sin(2\theta)) $$

$$ (x_2, y_2) = (-\sqrt{1 – \sin^2(2\theta)}, \sin(2\theta)) $$

Find the sine and cosine values at π/3

To find the sine and cosine values at $ \frac{\pi}{3} $:

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The unit circle values for $ \frac{\pi}{3} $ are:

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

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

Find the cosine of θ if sin(θ) = 1/2 and θ is in the first quadrant

Given $ \sin(\theta) = \frac{1}{2} $ and $ \theta $ is in the first quadrant.

We know that $ \sin^2(\theta) + \cos^2(\theta) = 1 $.

So,

$$ \left( \frac{1}{2} \right)^2 + \cos^2(\theta) = 1 $$

$$ \frac{1}{4} + \cos^2(\theta) = 1 $$

$$ \cos^2(\theta) = 1 – \frac{1}{4} $$

$$ \cos^2(\theta) = \frac{3}{4} $$

$$ \cos(\theta) = \pm \sqrt{\frac{3}{4}} $$

Since $ \theta $ is in the first quadrant, $ \cos(\theta) $ is positive:

$$ \cos(\theta) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $$

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:

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