Explore the unit circle and its relationship to angles, radians, trigonometric ratios, and coordinates in the coordinate plane.
Identify the location of -pi/2 on a unit circle
On the unit circle, angles are measured starting from the positive x-axis and moving counterclockwise.
To find the location of $-\pi/2$:
– Starting from the positive x-axis, move clockwise because the angle is negative.
– $\pi/2$ is 90 degrees, so moving clockwise 90 degrees lands you on the negative y-axis.
Therefore, the coordinates of $-\pi/2$ on the unit circle are:
$ (0, -1) $
Determine the value of theta for which the point (cos(θ), sin(θ)) on the unit circle forms a right-angled triangle with the origin and the point (1, 0)
Given the points $ (\cos(\theta), \sin(\theta))$, the origin $(0, 0)$, and $(1, 0)$, we need to find $ \theta $ such that they form a right-angled triangle.
The distance between $(\cos(\theta), \sin(\theta))$ and $(1, 0)$ is:
$$ d = \sqrt{(\cos(\theta) – 1)^2 + \sin^2(\theta)} $$
Since $(\cos(\theta), \sin(\theta))$ lies on the unit circle, we use the Pythagorean identity:
$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$
Thus the distance simplifies to:
$$ d = \sqrt{1 – 2\cos(\theta) + 1} = \sqrt{2 – 2\cos(\theta)} = \sqrt{2(1 – \cos(\theta))} $$
For the triangle to be right-angled, $\cos(\theta) = \frac{1}{2}$:
$$ \cos(\theta) = \frac{1}{2} \implies \theta = \frac{\pi}{3} \text{ or } \theta = -\frac{\pi}{3} $$
Calculate the area of a sector in a unit circle with a given central angle θ
To calculate the area of a sector in a unit circle with a given central angle $ \theta $ (in radians), use the following formula:
$$ A = \frac{1}{2} \cdot r^2 \cdot \theta $$
Since the radius $ r $ of a unit circle is 1, the formula simplifies to:
$$ A = \frac{1}{2} \cdot 1^2 \cdot \theta = \frac{\theta}{2} $$
So, the area of the sector is:
$$ A = \frac{\theta}{2} $$
Find the values of tan(θ), sin(θ), and cos(θ) for θ = 45 degrees
To find the values of $\tan(\theta)$, $\sin(\theta)$, and $\cos(\theta)$ for $\theta = 45^\circ$:
First, we note that $\theta = 45^\circ$ is in the first quadrant of the unit circle.
The coordinates of the point on the unit circle at $\theta = 45^\circ$ are:
$$\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$$
Therefore:
$$\sin(45^\circ) = \frac{\sqrt{2}}{2}$$
$$\cos(45^\circ) = \frac{\sqrt{2}}{2}$$
Using the definition of tangent:
$$\tan(45^\circ) = \frac{\sin(45^\circ)}{\cos(45^\circ)} = 1$$
Prove the identity involving cos and sin on the unit circle
To prove the identity involving $ \cos(\theta) $ and $ \sin(\theta) $ on the unit circle, we start with the Pythagorean identity:
$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$
Consider the parameterization of the unit circle with $ \theta $ as the angle from the positive x-axis:
$$ x = \cos(\theta) $$
$$ y = \sin(\theta) $$
Then, the coordinates $ (x, y) $ must satisfy:
$$ x^2 + y^2 = 1 $$
Substituting $ x = \cos(\theta) $ and $ y = \sin(\theta) $, we get:
$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$
This verifies the identity.
Find the coordinates of the point where the line y = 1 intersects the unit circle
To find the coordinates where the line $ y = 1 $ intersects the unit circle, we start by recalling the equation of the unit circle:
$$ x^2 + y^2 = 1 $$
Substituting $ y = 1 $ into the unit circle equation, we get:
$$ x^2 + 1^2 = 1 $$
Simplifying,
$$ x^2 + 1 = 1 $$
$$ x^2 = 0 $$
$$ x = 0 $$
Therefore, the point of intersection is:
$$ (0, 1) $$
Find the exact values of sin, cos, and tan at 30 degrees on the unit circle
First, we need to convert $ 30^{\circ} $ to radians:
$$ 30^{\circ} = \frac{\pi}{6} $$
Using the unit circle, the coordinates for $ \frac{\pi}{6} $ are $ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $
From this, we can find:
$$ \sin \frac{\pi}{6} = \frac{1}{2} $$
$$ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $$
$$ \tan \frac{\pi}{6} = \frac{\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$
Determine the value of tan(θ) when sin(θ) = 3/5 and θ is in the first quadrant
Given that $\sin(θ) = \frac{3}{5}$ and $θ$ is in the first quadrant, we can find $\cos(θ)$ using the Pythagorean identity:
$$\sin^2(θ) + \cos^2(θ) = 1$$
Plugging in the given value:
$$\left(\frac{3}{5}\right)^2 + \cos^2(θ) = 1$$
$$\frac{9}{25} + \cos^2(θ) = 1$$
$$\cos^2(θ) = 1 – \frac{9}{25} = \frac{16}{25}$$
Since $θ$ is in the first quadrant, $\cos(θ)$ is positive:
$$\cos(θ) = \frac{4}{5}$$
Now, we can find $\tan(θ)$:
$$\tan(θ) = \frac{\sin(θ)}{\cos(θ)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$$
Therefore, $\tan(θ) = \frac{3}{4}$.
Find the values of tan(θ) at various angles and verify using the unit circle
To find the values of $ \tan(\theta) $ at various angles and verify using the unit circle, we consider the following angles: $ \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} $
1. For $ \theta = \frac{\pi}{4} $:
$$ \tan(\frac{\pi}{4}) = 1 $$
2. For $ \theta = \frac{3\pi}{4} $:
$$ \tan(\frac{3\pi}{4}) = -1 $$
3. For $ \theta = \frac{5\pi}{4} $:
$$ \tan(\frac{5\pi}{4}) = 1 $$
4. For $ \theta = \frac{7\pi}{4} $:
$$ \tan(\frac{7\pi}{4}) = -1 $$
Verification: Using the unit circle, we observe that at these angles, the tangent value is consistent with the coordinates (x, y) where $ \tan(\theta) = \frac{y}{x} $.
Determine the coordinates of a point on the unit circle where the tangent line has a slope of 3/4
To determine the coordinates of a point on the unit circle where the tangent line has a slope of $\frac{3}{4}$, we start with the equation of the unit circle:
$$x^2 + y^2 = 1$$
The slope of the tangent line at a point $(x, y)$ on the circle can be found by differentiating implicitly:
$$2x + 2y\frac{dy}{dx} = 0$$
Solving for $\frac{dy}{dx}$, we get:
$$\frac{dy}{dx} = -\frac{x}{y}$$
We need the slope to equal $\frac{3}{4}$:
$$-\frac{x}{y} = \frac{3}{4}$$
This implies:
$$y = -\frac{4x}{3}$$
Substitute $y = -\frac{4x}{3}$ back into the unit circle equation:
$$x^2 + \left(-\frac{4x}{3}\right)^2 = 1$$
$$x^2 + \frac{16x^2}{9} = 1$$
$$\frac{25x^2}{9} = 1$$
$$x^2 = \frac{9}{25}$$
$$x = \pm \frac{3}{5}$$
Substitute $x$ back into $y = -\frac{4x}{3}$:
$$y = \mp \frac{4 \cdot \frac{3}{5}}{3} = \mp \frac{4}{5}$$
The coordinates are:
$$\left(\frac{3}{5}, -\frac{4}{5}\right)$$ and $$\left(-\frac{3}{5}, \frac{4}{5}\right)$$
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