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Find the angle θ in the unit circle such that cos(θ) = -1/2
We know that $ \cos(\theta) = -\frac{1}{2} $.
This value of cosine corresponds to two angles in the unit circle, which are in the second and third quadrants.
In the second quadrant, the reference angle is $ \theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $.
In the third quadrant, the reference angle is $ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $.
Therefore, $ \theta = \frac{2\pi}{3} $ or $ \theta = \frac{4\pi}{3} $.
Suppose that angle θ is positioned on the unit circle such that θ = 5π/6 Determine the coordinates of the point where the terminal side of θ intersects the unit circle
First, we identify that $ \theta = \frac{5\pi}{6} $ is in the second quadrant. The reference angle is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.
In the unit circle, the cosine and sine of $ \frac{\pi}{6} $ are $ \frac{\sqrt{3}}{2} $ and $ \frac{1}{2} $, respectively.
Therefore, in the second quadrant, the coordinates are $ (-\frac{\sqrt{3}}{2}, \frac{1}{2}) $.
$$ \text{Coordinates: } \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) $$
How to Find the Reference Angle Not on Unit Circle
To find the reference angle of an angle not on the unit circle, follow these steps:
1. Determine the quadrant in which the angle is located.
2. Use the following rules based on the quadrant to find the reference angle:
For an angle $\theta$ in the first quadrant, the reference angle is $\theta$.
For an angle $\theta$ in the second quadrant, the reference angle is $180^\circ – \theta$.
For an angle $\theta$ in the third quadrant, the reference angle is $\theta – 180^\circ$.
For an angle $\theta$ in the fourth quadrant, the reference angle is $360^\circ – \theta$.
Example: Find the reference angle for $210^\circ$.
Since $210^\circ$ is in the third quadrant, we use the rule for the third quadrant:
$$\text{Reference Angle} = 210^\circ – 180^\circ = 30^\circ$$
Therefore, the reference angle for $210^\circ$ is $30^\circ$.
Find the coordinates of a point on the unit circle at a given angle
To find the coordinates of a point on the unit circle at an angle of 120 degrees, we first convert the angle to radians, as the unit circle typically uses radians.
$$ \theta = 120^{\circ} = \frac{2\pi}{3} \text{ radians} $$
The coordinates on the unit circle are given by:
$$ (x, y) = (\cos \theta, \sin \theta) $$
Substituting our angle:
$$ x = \cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2} $$
$$ y = \sin \left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} $$
Thus, the coordinates of the point on the unit circle at 120 degrees are:
$$ \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) $$
Find the coordinates of the point on the unit circle at angle π/3 radians
To find the coordinates of the point on the unit circle at angle $\pi/3$ radians, we use the unit circle definitions. The unit circle is defined by the equation $x^2 + y^2 = 1$, where the coordinates $(x, y)$ correspond to $(\cos(\theta), \sin(\theta))$ for an angle $\theta$.
For $\theta = \pi/3$:
$$x = \cos(\pi/3) = \frac{1}{2}$$
$$y = \sin(\pi/3) = \frac{\sqrt{3}}{2}$$
Thus, the coordinates are $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$.
Find the intersection points of the unit circle and the line y = x – 1
To find the intersection points, we need to solve the system of equations formed by the unit circle equation and the given line equation.
The unit circle equation is:
$$x^2 + y^2 = 1$$
Substitute $y = x – 1$ into the unit circle equation:
$$x^2 + (x – 1)^2 = 1$$
Expand and simplify:
$$x^2 + x^2 – 2x + 1 = 1$$
Combine like terms:
$$2x^2 – 2x + 1 = 1$$
Simplify further:
$$2x^2 – 2x = 0$$
Factor out the common term:
$$2x(x – 1) = 0$$
Set each factor to zero:
$$x = 0$$
$$x = 1$$
For $x = 0$:
$$y = 0 – 1 = -1$$
For $x = 1$:
$$y = 1 – 1 = 0$$
Thus, the intersection points are $(0, -1)$ and $(1, 0)$.
Given the unit circle, if the point (a, b) lies on the circle, find the value of sin(2θ), cos(2θ), and tan(2θ) where θ is the angle that corresponds to the point (a, b) Verify that these values satisfy the double angle identities
Given the point (a, b) on the unit circle, we know:
$$a = \cos \theta $$
$$b = \sin \theta $$
We need to find $\sin(2\theta)$, $\cos(2\theta)$, and $\tan(2\theta)$. Using the double angle identities:
$$ \sin(2\theta) = 2 \sin \theta \cos \theta $$
$$ \cos(2\theta) = \cos^2 \theta – \sin^2 \theta $$
$$ \tan(2\theta) = \frac{2\tan \theta}{1 – \tan^2 \theta} $$
By substituting a = cos θ and b = sin θ:
$$ \sin(2\theta) = 2ab $$
$$ \cos(2\theta) = a^2 – b^2 $$
$$ \tan(2\theta) = \frac{2b/a}{1 – b^2/a^2} = \frac{2b/a}{(a^2 – b^2)/a^2} = \frac{2ab}{a^2 – b^2} $$
Verification of double-angle identities:
$$ (2ab)^2 + (a^2 – b^2)^2 = 4a^2b^2 + a^4 – 2a^2b^2 + b^4 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2 = 1 $$
Find the coordinates of point P on the unit circle at an angle of 45 degrees
To find the coordinates of point P on the unit circle at an angle of 45 degrees, we use the fact that the unit circle has a radius of 1 and that the coordinates correspond to the cosine and sine of the angle.
$$\cos(45^\circ) = \frac{\sqrt{2}}{2}$$
$$\sin(45^\circ) = \frac{\sqrt{2}}{2}$$
Therefore, the coordinates of point P are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
Find the exact coordinates of a point on the unit circle given that the point is 7π/6 radians from the positive x-axis
To determine the coordinates of the point on the unit circle at an angle of $\frac{7\pi}{6}$ radians, we use the sine and cosine functions:
The x-coordinate (cosine) is:
$$\cos\left(\frac{7\pi}{6}\right) = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$
The y-coordinate (sine) is:
$$\sin\left(\frac{7\pi}{6}\right) = \sin\left(\pi + \frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$$
Thus, the exact coordinates are:
$$\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$$
Find the sine and cosine values for the angle 30 degrees on the unit circle
To find the sine and cosine values for the angle $30^{\circ}$ on the unit circle, we use the known values of sine and cosine for common angles. The coordinates of the point on the unit circle at $30^{\circ}$ are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$.
Therefore,
$$ \sin(30^{\circ}) = \frac{1}{2} $$
$$ \cos(30^{\circ}) = \frac{\sqrt{3}}{2} $$
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