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Identify the quadrant of the angle theta = pi/3
To determine the quadrant of the angle $ \theta = \frac{\pi}{3} $, we convert it to degrees:
$$ \theta = \frac{\pi}{3} \times \frac{180}{\pi} = 60^{\circ} $$
The angle $ 60^{\circ} $ lies in the first quadrant.
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 coordinates of points on the unit circle corresponding to specific angles
To find the coordinates of points on the unit circle corresponding to $ \theta = \frac{\pi}{6}, \theta = \frac{\pi}{4}, \theta = \frac{\pi}{3} $, we use the unit circle properties:
For $ \theta = \frac{\pi}{6} $:
$$ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $$
For $ \theta = \frac{\pi}{4} $:
$$ (\cos(\frac{\pi}{4}), \sin(\frac{\pi}{4})) = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) $$
For $ \theta = \frac{\pi}{3} $:
$$ (\cos(\frac{\pi}{3}), \sin(\frac{\pi}{3})) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $$
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:
$$ (1, 0), (-1, 0), (1, 0) $$
Determine the coordinates of the points on the unit circle where the angle is pi/4
To determine the coordinates of the points on the unit circle where the angle is $ \frac{\pi}{4} $, we need to use trigonometric functions.
On the unit circle, the x-coordinate is given by $ \cos(\theta) $ and the y-coordinate is given by $ \sin(\theta) $, where $ \theta $ is the angle.
For $ \theta = \frac{\pi}{4} $:
$$ x = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$
$$ y = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$
Thus, the coordinates are:
$$ \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) $$
Find the reference angle for a given angle of 345 degrees in the unit circle
To find the reference angle for $345^\circ$, note that it is in the fourth quadrant. The reference angle in the fourth quadrant is found by subtracting the given angle from $360^\circ$:
$$ 360^\circ – 345^\circ = 15^\circ $$
So, the reference angle for $345^\circ$ is:
$$ 15^\circ $$
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 tangent line equations for every point on the unit circle
To find the tangent line equations for every point on the unit circle, we start with the unit circle equation:
$$ x^2 + y^2 = 1 $$
Differentiate implicitly with respect to $x$ to find the slope:
$$ 2x + 2y \x0crac{dy}{dx} = 0 $$
Solve for $ \x0crac{dy}{dx} $:
$$ \x0crac{dy}{dx} = -\x0crac{x}{y} $$
At a point $ (a, b) $ on the unit circle, the slope of the tangent is:
$$ m = -\x0crac{a}{b} $$
The tangent line equation at $ (a, b) $ is:
$$ y – b = -\x0crac{a}{b}(x – a) $$
Multiply through by $ b $ to get:
$$ b(y – b) = -a(x – a) $$
Simplify to obtain the final equation of the tangent line:
$$ ax + by = 1 $$
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