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Find the point on the unit circle where the sine value is negative and the cosine value is positive
The unit circle is defined as the set of all points $(x, y)$ such that:
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$$ x^2 + y^2 = 1 $$
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In the unit circle, the sine value corresponds to the y-coordinate and the cosine value corresponds to the x-coordinate. We need to find a point where:
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$$ y < 0 $$
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$$ x > 0 $$
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One such point is:
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$$ \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $$
Determine the coordinates of points on the unit circle that satisfy the given equation
To determine the coordinates of points on the unit circle that satisfy the equation $ \cos^2(\theta) – \sin^2(\theta) = 0 $:
First, we recall the Pythagorean identity: $$ \cos^2(\theta) + \sin^2(\theta) = 1 $$
Given the equation: $$ \cos^2(\theta) – \sin^2(\theta) = 0 $$
This can be rewritten as: $$ \cos^2(\theta) = \sin^2(\theta) $$
Taking the square root of both sides gives: $$ \cos(\theta) = \pm \sin(\theta) $$
We consider the positive and negative cases separately.
For $ \cos(\theta) = \sin(\theta) $:
$$ \theta = \frac{\pi}{4} + k \pi $$
Where $ k $ is any integer.
For $ \cos(\theta) = -\sin(\theta) $:
$$ \theta = \frac{3\pi}{4} + k \pi $$
Where $ k $ is any integer.
Thus, the coordinates of the points on the unit circle are:
$$ ( \cos(\frac{\pi}{4} + k \pi), \sin(\frac{\pi}{4} + k \pi) ) $$
$$ ( \cos(\frac{3\pi}{4} + k \pi), \sin(\frac{3\pi}{4} + k \pi) ) $$
What is the value of cos(π/4) on the unit circle?
Find the coordinates of the point on the unit circle where the inverse sine function is equal to 1/2
To find the coordinates of the point on the unit circle where the inverse sine function, $ \sin^{-1}(x) $, is equal to $ \frac{1}{2} $:
We need to solve the equation:
$$ \sin^{-1}(y) = \frac{1}{2} $$
The angle whose sine is $ \frac{1}{2} $ is:
$$ \theta = \frac{\pi}{6} $$
On the unit circle, the coordinates corresponding to this angle are:
$$ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) $$
Evaluating the trigonometric functions, we get:
$$ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $$
$$ \sin(\frac{\pi}{6}) = \frac{1}{2} $$
So, the coordinates are:
$$ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $$
Determine the value of sec(θ) given cos(θ) = -1/sqrt(2) and θ is in the third quadrant
Given that $\cos(\theta) = -\frac{1}{\sqrt{2}}$ and $\theta$ is in the third quadrant, we start by using the identity:
$$\sec(\theta) = \frac{1}{\cos(\theta)}$$
Substituting the given value:
$$\sec(\theta) = \frac{1}{-\frac{1}{\sqrt{2}}}$$
We simplify the fraction:
$$\sec(\theta) = -\sqrt{2}$$
Thus, the value of $\sec(\theta)$ is $-\sqrt{2}$.
Find the sine and cosine values at t = pi/4 on the unit circle
To find the sine and cosine values at $ t = \frac{\pi}{4} $ on the unit circle, we use the following values:
$$ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $$
$$ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $$
Thus, the sine and cosine values are:
$$ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $$
Identify the coordinates on the unit circle for θ = π/4
To find the coordinates on the unit circle for $ \theta = \frac{\pi}{4} $, we use the unit circle properties.
For $ \theta = \frac{\pi}{4} $, the coordinates are given by:
$$ (\cos(\frac{\pi}{4}), \sin(\frac{\pi}{4})) $$
From trigonometric values, we know:
$$ \cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$
So the coordinates are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
Find the values of sin and cos at the given angle on the unit circle
To find the values of $\sin$ and $\cos$ at an angle of $\theta = \frac{\pi}{3}$ on the unit circle:
$$\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$$
$$\cos(\frac{\pi}{3}) = \frac{1}{2}$$
Find the angle in radians for a point on the unit circle where the x-coordinate is 1/2
To find the angle $\theta$ in radians for a point on the unit circle where the $x$-coordinate is $\frac{1}{2}$, we consider the cosine function:
$$\cos(\theta) = \frac{1}{2}$$
The angles that satisfy this equation are:
$$\theta = \frac{\pi}{3}$$ and $$\theta = -\frac{\pi}{3}$$ or equivalently $$\theta = \frac{5\pi}{3}$$
What is the value of sin(15 degrees) using the unit circle?
To find the value of $ \sin(15^\circ) $ using the unit circle, we use the angle addition formula:
$$ \sin(a + b) = \sin(a) \cos(b) + \cos(a)\sin(b) $$
Here, let $ a = 45^\circ $ and $ b = -30^\circ $. Then,
$$ \sin(45^\circ – 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) $$
Substituting the values, we get:
$$ \sin(15^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} $$
Simplify the expression:
$$ \sin(15^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} $$
Combine the fractions:
$$ \sin(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} $$
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