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Identify the coordinates on the unit circle for angle θ
The coordinates of a point on the unit circle for a given angle $ \theta $ are given by:
$$ ( \cos(\theta), \sin(\theta) ) $$
For example, if $ \theta = \frac{\pi}{4} $:
$$ ( \cos(\frac{\pi}{4}), \sin(\frac{\pi}{4}) ) = ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ) $$
Find the values of tan(θ) for θ in the interval [0, 2π] that satisfy the equation tan(θ) = 2
To find the values of $ \tan(\theta) $ that satisfy the equation $ \tan(\theta) = 2 $ in the interval $ [0, 2\pi] $, we need to determine the angles where the tangent function equals 2.
First, recall that the tangent function is periodic with period $ \pi $, and the angles where $ \tan(\theta) = 2 $ are:
$$ \theta_1 = \arctan(2) $$
and
$$ \theta_2 = \arctan(2) + \pi $$
Because the tangent function repeats every $ \pi $ radians, we only need to check within one period:
$$ \theta_1 = \arctan(2) $$
$$ \theta_2 = \arctan(2) + \pi $$
Thus, the solutions within $ [0, 2\pi] $ are:
$$ \theta = \arctan(2) $$
and
$$ \theta = \arctan(2) + \pi $$
Solve the equation to find all distinct points where the ellipse intersects the unit circle
To find where the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intersects the unit circle $x^2 + y^2 = 1$, we need to solve the system of equations:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
$$ x^2 + y^2 = 1 $$
First, substitute $y^2 = 1 – x^2$ into the ellipse equation:
$$ \frac{x^2}{a^2} + \frac{1 – x^2}{b^2} = 1 $$
Multiply through by $a^2b^2$ to clear the denominators:
$$ b^2x^2 + a^2(1 – x^2) = a^2b^2 $$
Simplify and solve for $x^2$:
$$ b^2x^2 + a^2 – a^2x^2 = a^2b^2 $$
$$ (b^2 – a^2)x^2 = a^2(b^2 – 1) $$
$$ x^2 = \frac{a^2(b^2 – 1)}{b^2 – a^2} $$
Then solve for $y^2$ using $y^2 = 1 – x^2$.
Determine the sine and cosine values of 5π/6
To determine the sine and cosine values of $ \frac{5\pi}{6} $, we refer to the unit circle.
The angle $ \frac{5\pi}{6} $ is located in the second quadrant.
In the second quadrant, sine is positive, and cosine is negative.
The reference angle for $ \frac{5\pi}{6} $ is $ \frac{\pi}{6} $.
Therefore:
$$ \sin(\frac{5\pi}{6}) = \sin(\pi – \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} $$
$$ \cos(\frac{5\pi}{6}) = \cos(\pi – \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} $$
Find the sine and cosine of the angle π/3 using the unit circle
To find the sine and cosine of the angle $ \pi/3 $ using the unit circle, consider the angle that corresponds to $ \pi/3 $ radians (or 60 degrees).
In the unit circle, the coordinates of the point on the circumference corresponding to the angle $ \pi/3 $ are $ (\cos(\pi/3), \sin(\pi/3)) $.
For $ \pi/3 $:
$$ \cos(\pi/3) = \frac{1}{2} $$
$$ \sin(\pi/3) = \frac{\sqrt{3}}{2} $$
Find the coordinates of the point where the terminal side of the angle intersects the unit circle at an angle of 5π/4 radians
To find the coordinates of the point where the terminal side of the angle intersects the unit circle at an angle of $\frac{5\pi}{4}$ radians, we use the unit circle properties.
The angle $\frac{5\pi}{4}$ radians is in the third quadrant where both sine and cosine values are negative.
The reference angle for $\frac{5\pi}{4}$ is $\frac{\pi}{4}$.
The coordinates on the unit circle for an angle of $\frac{\pi}{4}$ are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.
Since we are in the third quadrant, we change the signs of both x and y coordinates:
Therefore, the coordinates are:
$$ \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $$
Find the coordinates of cos(π/3) on the unit circle
To find the coordinates of $ \cos(\frac{\pi}{3}) $ on the unit circle, we need to identify the coordinates associated with this angle.
On the unit circle, the angle $ \frac{\pi}{3} $ corresponds to the 60° position.
At this position, the coordinates are:
$$ ( \cos(\frac{\pi}{3}), \sin(\frac{\pi}{3}) ) = ( \frac{1}{2}, \frac{\sqrt{3}}{2} ) $$
Determine the general solution for sin(x) = 1/2 within [0, 2π]
To determine the general solution for $ \sin(x) = \frac{1}{2} $ within the interval $ [0, 2\pi] $, we need to find all the angles $ x $ where the sine function yields $ \frac{1}{2} $ on the unit circle.
The sine function equals $ \frac{1}{2} $ at angles $ \frac{\pi}{6} $ and $ \frac{5\pi}{6} $ within the given interval.
Thus, the general solutions are:
$$ x = \frac{\pi}{6} + 2n\pi $$
and
$$ x = \frac{5\pi}{6} + 2n\pi $$
where $ n $ is any integer.
Determine the coordinates of points on the unit circle where the tangent line is horizontal
To find the coordinates on the unit circle where the tangent line is horizontal, we first recall that the unit circle is defined by the equation:
$$ x^2 + y^2 = 1 $$
The slope of the tangent line to the circle at any point (x, y) is given by the derivative of y with respect to x. Differentiating implicitly, we get:
$$ 2x + 2y \x0crac{dy}{dx} = 0 $$
Solving for $\x0crac{dy}{dx}$, we find:
$$ \x0crac{dy}{dx} = -\x0crac{x}{y} $$
For the tangent line to be horizontal, the slope $\x0crac{dy}{dx}$ must be zero. This occurs when:
$$ -\x0crac{x}{y} = 0 $$
Thus, x must be zero. On the unit circle, the points with x = 0 are (0, 1) and (0, -1). Therefore, the coordinates are (0, 1) and (0, -1).
What are the coordinates of 3π/4 on the unit circle?
The coordinates of $ \frac{3\pi}{4} $ on the unit circle can be found using the unit circle definitions. The angle $ \frac{3\pi}{4} $ corresponds to $ 135^{\circ} $. At this angle, the coordinates are:
$$ \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
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