Explore the unit circle and its relationship to angles, radians, trigonometric ratios, and coordinates in the coordinate plane.
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:
$$ ax + by = 1 $$
Find the value of sec(θ) for θ = π/4 on the unit circle
To find the value of $ \sec(\theta) $ for $ \theta = \frac{\pi}{4} $ on the unit circle, we use the definition of secant, which is the reciprocal of cosine:
$$ \sec(\theta) = \frac{1}{\cos(\theta)} $$
For $ \theta = \frac{\pi}{4} $, we have:
$$ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
Therefore:
$$ \sec\left(\frac{\pi}{4}\right) = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} $$
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:
\n
$$ x^2 + y^2 = 1 $$
\n
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:
\n
$$ y < 0 $$
\n
$$ x > 0 $$
\n
One such point is:
\n
$$ \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $$
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:
$$ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $$
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