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Find the Cosine of 45 Degrees
To find the cosine of $45^\circ$, we use the unit circle. On the unit circle, the coordinates of the point where the terminal side of the $45^\circ$ angle intersects the circle are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$. The cosine of an angle is equal to the x-coordinate of this point.
Therefore,
$$\cos(45^\circ) = \frac{\sqrt{2}}{2}$$
Problem: Calculate the Sine, Cosine, and Tangent Values of Specific Angles on the Unit Circle
Let’s determine the sine, cosine, and tangent values for the angle θ = 225° on the unit circle.
First, convert the angle to radians:
$$ θ = 225° = \frac{225π}{180} = \frac{5π}{4} radians $$
Using the properties of the unit circle, we know:
$$ \cos(\frac{5π}{4}) = -\frac{\sqrt{2}}{2} $$
$$ \sin(\frac{5π}{4}) = -\frac{\sqrt{2}}{2} $$
$$ \tan(\frac{5π}{4}) = \frac{\sin(\frac{5π}{4})}{\cos(\frac{5π}{4})} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 $$
Thus, the sine, cosine, and tangent values for θ = 225° are:
$$ \sin(225°) = -\frac{\sqrt{2}}{2} $$
$$ \cos(225°) = -\frac{\sqrt{2}}{2} $$
$$ \tan(225°) = 1 $$
Find the length of the chord in a unit circle
Consider a unit circle with center at the origin (0,0). Let the endpoints of the chord be at coordinates (cos θ, sin θ) and (cos φ, sin φ).
The formula for finding the distance between two points (x1, y1) and (x2, y2) is given by:
$$ d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} $$
For points (cos θ, sin θ) and (cos φ, sin φ), the distance (chord length) is:
$$ d = \sqrt{(\cos φ – \cos θ)^2 + (\sin φ – \sin θ)^2} $$
Using trigonometric identities, we get:
$$ d = \sqrt{2 – 2 \cos(θ – φ)} $$
Given that this is a unit circle, we simplify as:
$$ d = 2 \sin \left(\frac{θ – φ}{2}\right) $$
Therefore, the length of the chord is:
$$ 2 \sin \left(\frac{θ – φ}{2}\right) $$
Calculating the Tangent Value of an Angle in the Unit Circle
Let’s consider an angle $ \theta $ in the unit circle. The coordinates of a point on the unit circle are given by $(\cos \theta, \sin \theta)$. The tangent of the angle $ \theta $ is defined as:
$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$
Suppose $\theta = \frac{5\pi}{4}$, we need to find the value of $\tan \theta$. From the unit circle, we have:
$\sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$
$\cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$
Thus,
$$\tan \frac{5\pi}{4} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1$$
How to Find the Tangent of a Point on the Unit Circle
Given a point on the unit circle, say $(\cos(\theta), \sin(\theta))$, we need to find the tangent line at this point.
Step 1: The equation of the unit circle is $x^2 + y^2 = 1$.
Step 2: To find the slope of the tangent, we differentiate implicitly with respect to $x$:
$$2x + 2y \frac{dy}{dx} = 0$$
Step 3: Rearrange to solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = -\frac{x}{y}$$
Step 4: Substitute $x = \cos(\theta)$ and $y = \sin(\theta)$:
$$\frac{dy}{dx} = -\frac{\cos(\theta)}{\sin(\theta)} = -\cot(\theta)$$
So the slope of the tangent line at $(\cos(\theta), \sin(\theta))$ is $-\cot(\theta)$.
Determine the tangent values for specific angles on the unit circle
To determine the tangent values for angles $\frac{\pi}{4}$, $\frac{2\pi}{3}$, and $\frac{5\pi}{6}$ on the unit circle, follow these steps:
1. For the angle $\frac{\pi}{4}$: $$\tan \left( \frac{\pi}{4} \right) = 1$$
2. For the angle $\frac{2\pi}{3}$: $$\tan \left( \frac{2\pi}{3} \right) = -\sqrt{3}$$
3. For the angle $\frac{5\pi}{6}$: $$\tan \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{3}$$
Thus, the tangent values are $1$, $-\sqrt{3}$, and $-\frac{\sqrt{3}}{3}$ respectively.
Given that the angle θ in standard position intersects the unit circle at the point (x, y) in the first quadrant where x = 3/5, find the y-coordinate of the point Use the Pythagorean identity for the unit circle to show your work
Given the Pythagorean identity for the unit circle:
$$ x^2 + y^2 = 1 $$
where $$ x = \frac{3}{5}$$, substitute this value into the identity:
$$ \left( \frac{3}{5} \right)^2 + y^2 = 1 $$
$$ \frac{9}{25} + y^2 = 1 $$
Subtract $$ \frac{9}{25}$$ from both sides:
$$ y^2 = 1 – \frac{9}{25} $$
$$ y^2 = \frac{25}{25} – \frac{9}{25} $$
$$ y^2 = \frac{16}{25} $$
Taking the square root of both sides:
$$ y = \pm \sqrt{\frac{16}{25}} $$
$$ y = \pm \frac{4}{5} $$
Since (x, y) is in the first quadrant:
$$ y = \frac{4}{5} $$
Find the angle whose cosine is -2/3 using the unit circle
To find the angle whose cosine is $-\frac{2}{3}$, we need to look at the unit circle and identify the angles where the x-coordinate (cosine value) is $-\frac{2}{3}$. Since cosine is negative in the second and third quadrants, we look in those regions.
Thus, we have:
$$\theta = \cos^{-1}(-\frac{2}{3})$$
and
$$\theta = 2\pi – \cos^{-1}(-\frac{2}{3})$$
These angles in degrees are approximately:
$$\theta \approx 131.81^\circ$$
and
$$\theta \approx 228.19^\circ$$
Find the Cotangent of an Angle on the Unit Circle
To find the cotangent of an angle $\theta$ on the unit circle, we use the identity:
$$ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} $$
Given $\theta = \frac{3\pi}{4}$, we know from the unit circle that:
$$ \cos \left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$
and
$$ \sin \left( \frac{3\pi}{4} \right) = \frac{\sqrt{2}}{2} $$
Therefore,
$$ \cot \left( \frac{3\pi}{4} \right) = \frac{\cos \left( \frac{3\pi}{4} \right)}{\sin \left( \frac{3\pi}{4} \right)} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$
So, the cotangent of $\frac{3\pi}{4}$ is $-1$.
Find the angle where tan(θ) = -1 in the unit circle
To find the angle where $\tan(\theta) = -1$ in the unit circle, we need to look for the values of $\theta$ where the tangent function is negative and equals -1.
We know that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. For $\tan(\theta) = -1$, this means $\sin(\theta) = -\cos(\theta)$.
This occurs in the second and fourth quadrants.
In the second quadrant: $\theta = \pi – \frac{\pi}{4} = \frac{3\pi}{4}$
In the fourth quadrant: $\theta = 2\pi – \frac{\pi}{4} = \frac{7\pi}{4}$
Hence, the angles are $\theta = \frac{3\pi}{4}$ and $\theta = \frac{7\pi}{4}$.
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