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Find the secant value of angle π/3 on the unit circle
To find the secant value, we first need to know the cosine value of the given angle on the unit circle.
The angle $\frac{\pi}{3}$ corresponds to an angle of $60^\circ$.
On the unit circle, the cosine of $\frac{\pi}{3}$ is $\frac{1}{2}$.
The secant is the reciprocal of the cosine:
$$\sec\left(\frac{\pi}{3}\right) = \frac{1}{\cos\left(\frac{\pi}{3}\right)} = \frac{1}{\frac{1}{2}} = 2$$
Given an angle of 5π/6 radians, find the coordinates of the point on the unit circle corresponding to this angle
Given an angle of $\frac{5\pi}{6}$ radians, we need to find the coordinates of the point on the unit circle corresponding to this angle.
First, note that $\frac{5\pi}{6}$ radians lies in the second quadrant. The reference angle for $\frac{5\pi}{6}$ is $\pi – \frac{5\pi}{6} = \frac{\pi}{6}$.
The coordinates of the point corresponding to $\frac{\pi}{6}$ on the unit circle are $(\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) = (\frac{\sqrt{3}}{2}, \frac{1}{2})$.
Since $\frac{5\pi}{6}$ is in the second quadrant, the x-coordinate is negative and the y-coordinate is positive. Thus, the coordinates of the point are $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.
Find the sine of the given angle on the unit circle
Given the angle $ \theta = 30^{\circ} $, we need to find $ \sin(\theta) $ using the unit circle.
On the unit circle, the coordinates of the point corresponding to $ 30^{\circ} $ are $ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $. The sine of an angle is the y-coordinate of this point. Therefore,
$$ \sin(30^{\circ}) = \frac{1}{2} $$
Understanding the Unit Circle: An Advanced Problem
To understand the unit circle at an advanced level, consider the problem of determining the exact value of trigonometric functions given a point on the unit circle. Suppose a point $P$ on the unit circle corresponds to an angle $\theta$. Given $P = \left(-\frac{3}{5}, -\frac{4}{5}\right)$, find $\sin \theta$, $\cos \theta$, $\tan \theta$, and the corresponding coordinates for $\theta + 2\pi$.
First, recall that for any point $(x, y)$ on the unit circle:
$$ x = \cos \theta, \quad y = \sin \theta $$
Thus, for $P = \left(-\frac{3}{5}, -\frac{4}{5}\right)$, we have:
$$ \cos \theta = -\frac{3}{5}, \quad \sin \theta = -\frac{4}{5} $$
Next, compute $\tan \theta$:
$$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} $$
Lastly, for the angle $\theta + 2\pi$, the coordinates remain the same since $2\pi$ represents a full rotation around the unit circle:
$$ P_{\theta + 2\pi} = \left(-\frac{3}{5}, -\frac{4}{5}\right) $$
Determine the cosine and sine values for an angle of 45 degrees on the unit circle
To determine the $\cos$ and $\sin$ values for an angle of $45^\circ$ on the unit circle, follow these steps:
1. Convert the angle from degrees to radians: $45^\circ = \frac{\pi}{4}$ radians.
2. On the unit circle, the coordinates of a point corresponding to an angle of $\frac{\pi}{4}$ radians are given by $(\cos(\frac{\pi}{4}), \sin(\frac{\pi}{4}))$.
3. Using trigonometric values, we know:
$$\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$
$$\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$
Thus, the cosine and sine values for an angle of $45^\circ$ are $\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{2}}{2}$ respectively.
Determining the Position of -pi/2 on a Unit Circle
First, we recognize that the unit circle is a circle with a radius of 1 centered at the origin (0, 0) in the Cartesian coordinate system. The angle \(-\frac{\pi}{2}\) radians corresponds to a rotation in the clockwise direction from the positive x-axis.
Since \(\frac{\pi}{2}\) radians corresponds to 90 degrees, \(-\frac{\pi}{2}\) represents a rotation of 90 degrees clockwise. On the unit circle, rotating 90 degrees clockwise from the positive x-axis brings us to the negative y-axis.
Therefore, the coordinates on the unit circle at \(-\frac{\pi}{2}\) are:
$$ (0, -1) $$
Determine the values of trigonometric functions using the unit circle
To find the exact values of the trigonometric functions for the angle $ \theta = \frac{5\pi}{4} $ using the unit circle, follow these steps:
1. Locate the angle $ \theta = \frac{5\pi}{4} $ on the unit circle. This angle corresponds to $ 225^{\circ} $, or $ 45^{\circ} $ in the third quadrant.
2. In the third quadrant, both the sine and cosine values are negative. The reference angle is $ 45^{\circ} $.
3. The coordinates for $ 45^{\circ} $ are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $, so for $ 225^{\circ} $ these coordinates are $ \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $.
4. Therefore, $ \sin\left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $ and $ \cos\left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $.
5. The tangent function is $ \tan\left( \frac{5\pi}{4} \right) = \frac{\sin\left( \frac{5\pi}{4} \right)}{\cos\left( \frac{5\pi}{4} \right)} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 $.
Unit Circle and Trigonometric Functions
Consider the unit circle centered at the origin. Let point $P$ have coordinates $(x, y)$ on the unit circle such that the angle $\theta$ formed by the positive x-axis and the radius to $P$ is in the fourth quadrant. If the secant and tangent of $\theta$ are given as $\sec \theta = 5$ and $\tan \theta = -\frac{4}{3}$, find the coordinates of point $P$.
Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\cos \theta = \frac{1}{5}$. Since $x^2 + y^2 = 1$ for any point on the unit circle:
$$x^2 + y^2 = 1$$
Given that $x = \cos \theta = \frac{1}{5}$, we find $y$ using $\tan \theta = \frac{y}{x}$:
$$y = \tan \theta \cdot x = -\frac{4}{3} \cdot \frac{1}{5} = -\frac{4}{15}$$
Therefore, the coordinates of point $P$ are:
$$P = \left( \frac{1}{5}, -\frac{4}{15} \right)$$
Determine the coordinates of a point on the flipped unit circle given certain conditions
Let’s consider the unit circle equation flipped along y=x: $$x^2 + y^2 = 1$$ becomes $$y = x \cdot \sqrt{1 – x^2}$$. Given a point where the x-coordinate is $$\frac{1}{2}$$, find the corresponding y-coordinate.
Since the point lies on the flipped unit circle, we have:
$$y = \frac{1}{2} \cdot \sqrt{1 – (\frac{1}{2})^2}$$
$$y = \frac{1}{2} \cdot \sqrt{1 – \frac{1}{4}}$$
$$y = \frac{1}{2} \cdot \sqrt{\frac{3}{4}}$$
$$y = \frac{1}{2} \cdot \frac{\sqrt{3}}{2}$$
$$y = \frac{\sqrt{3}}{4}$$
Hence, the point on the flipped unit circle is $$\left(\frac{1}{2}, \frac{\sqrt{3}}{4}\right)$$.
Determine the exact values of sine, cosine, and tangent for a given angle on the unit circle
Let $\theta$ be an angle on the unit circle such that $\theta = \frac{7\pi}{6}$. Determine the exact values of $\sin(\theta)$, $\cos(\theta)$, and $\tan(\theta)$.
The angle $\theta = \frac{7\pi}{6}$ is in the third quadrant. The reference angle is $\pi – \frac{\pi}{6} = \frac{\pi}{6}$. In the third quadrant, both sine and cosine are negative.
Therefore, $\sin\left(\frac{7\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$
$\cos\left(\frac{7\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
and $\tan\left(\frac{7\pi}{6}\right) = \frac{\sin\left(\frac{7\pi}{6}\right)}{\cos\left(\frac{7\pi}{6}\right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
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