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Find all the solutions for cos(θ) = -1/2 on the unit circle
$$ \text{We need to find all } \theta \text{ such that } \cos(\theta) = -\frac{1}{2}. $$
$$ \text{The values of } \theta \text{ where } \cos(\theta) = -\frac{1}{2} \text{ are at } \theta = \frac{2\pi}{3} + 2k\pi \text{ and } \theta = \frac{4\pi}{3} + 2k\pi \text{ for any integer } k. $$
$$ \text{Thus, all solutions are: } \theta = \frac{2\pi}{3} + 2k\pi \text{ and } \theta = \frac{4\pi}{3} + 2k\pi. $$
Find the value of cotangent for an angle of 45 degrees on the unit circle
To find $\cot(45^\circ)$ on the unit circle, we start by recalling that cotangent is the reciprocal of the tangent function.
Given:
$$\cot(\theta) = \frac{1}{\tan(\theta)} $$
For $\theta = 45^\circ$:
$$\tan(45^\circ) = 1 $$
Thus,
$$\cot(45^\circ) = \frac{1}{1} = 1 $$
Hence, the value of $\cot(45^\circ)$ is 1.
Find the Quadrant of a Given Angle on the Unit Circle
To determine the quadrant in which the angle 150° lies, we first convert it to radians:
$$150° \times \frac{\pi}{180°} = \frac{5\pi}{6}$$
The angle \(\frac{5\pi}{6}\) is greater than \(\frac{\pi}{2}\) but less than \(\pi\). Hence, it lies in the second quadrant.
Find the sine and cosine of the angle at three specific points on the unit circle
The three specific points we will consider are $\frac{\pi}{6}$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$.
1. For $\frac{\pi}{6}$:
The sine value can be found using the unit circle as $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
The cosine value can be found as $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.
2. For $\frac{\pi}{4}$:
The sine value can be found as $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.
The cosine value can be found as $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.
3. For $\frac{\pi}{3}$:
The sine value can be found as $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.
The cosine value can be found as $\cos(\frac{\pi}{3}) = \frac{1}{2}$.
Therefore, the values are:
$\sin(\frac{\pi}{6}) = \frac{1}{2}$, $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$
$\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$
$\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, $\cos(\frac{\pi}{3}) = \frac{1}{2}$
Given a point on the unit circle at coordinates (sqrt(3)/2, -1/2), find the angle in radians, the corresponding angle in degrees, and the sine of the angle
To find the angle in radians, we use the coordinates on the unit circle. The x-coordinate gives us $$\cos(\theta) = \frac{\sqrt{3}}{2}$$ and the y-coordinate gives us $$\sin(\theta) = -\frac{1}{2}$$. These coordinates correspond to an angle in the fourth quadrant. Therefore, the angle in radians is:
$$\theta = -\frac{\pi}{6}$$
To convert this to degrees, we use the conversion factor $$\frac{180}{\pi}$$:
$$\theta = -\frac{\pi}{6} \times \frac{180}{\pi} = -30^\circ$$
The sine of the angle is already given by the y-coordinate:
$$\sin(\theta) = -\frac{1}{2}$$
Find the sine, cosine, and tangent of π/6 on the unit circle
To find the sine, cosine, and tangent of $\frac{\pi}{6}$, we use the unit circle values:
Sine of $\frac{\pi}{6}$: $$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$
Cosine of $\frac{\pi}{6}$: $$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$
Tangent of $\frac{\pi}{6}$: $$\tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$
Given a point (x,y) on the unit circle where the angle in radians is \(\theta\), express the coordinates of the point in terms of \(\cos(\theta)\) and \(\sin(\theta)\), and show how these expressions can be derived from the Pythagorean identity Additiona
The unit circle is defined by the equation:
$$ x^2 + y^2 = 1 $$
Since the point \((x,y)\) lies on the unit circle, we can express \(x\) and \(y\) in terms of \(\cos(\theta)\) and \(\sin(\theta)\):
$$ x = \cos(\theta) $$
$$ y = \sin(\theta) $$
These expressions satisfy the Pythagorean identity:
$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$
Now, given \( \cos(\theta) = -\frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \), we need to find the angle \(\theta\).
The values correspond to the angle \( \theta = \frac{2\pi}{3} \) or \( \theta = \frac{4\pi}{3} \) (in the second and third quadrants respectively where cosine is negative and sine is positive).
Therefore, the angles in radians are:
$$ \theta = \frac{2\pi}{3}, \frac{4\pi}{3} $$
Find the value of sin, cos, and tan for 45 degrees using the unit circle
To find the values of $\sin$, $\cos$, and $\tan$ for $45^\circ$ using the unit circle, we first note that $45^\circ$ is equivalent to $\frac{\pi}{4}$ radians.
On the unit circle, the coordinates for $\frac{\pi}{4}$ radians (or $45^\circ$) are $( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} )$.
Therefore:
$$\sin(45^\circ) = \frac{\sqrt{2}}{2} $$
$$\cos(45^\circ) = \frac{\sqrt{2}}{2} $$
$$\tan(45^\circ) = \frac{\sin(45^\circ)}{\cos(45^\circ)} = 1 $$
Consider a unit circle with center at the origin A square is inscribed in the circle, and a triangle is inscribed in the square Determine the area of the triangle
$$\text{First, determine the side length of the square inscribed in the unit circle. The diagonal of the square is equal to the diameter of the circle, which is 2.}$$
$$\text{Using Pythagoras’ theorem, the side length } a \text{ of the square is given by:}$$
$$a\sqrt{2} = 2$$
$$a = \frac{2}{\sqrt{2}} = \sqrt{2}$$
$$\text{Next, consider an equilateral triangle inscribed in the square. The side length of the triangle is the same as the side length of the square, } a = \sqrt{2}.\text{ The area of an equilateral triangle with side length } a \text{ is given by: }$$
$$A = \frac{\sqrt{3}}{4} a^2$$
$$A = \frac{\sqrt{3}}{4} (\sqrt{2})^2 = \frac{\sqrt{3}}{4} \times 2 = \frac{\sqrt{3}}{2}$$
$$\text{Therefore, the area of the triangle is } \frac{\sqrt{3}}{2}. $$
Find the coordinates on the unit circle at different angles
To find the coordinates on the unit circle at 45 degrees:
The unit circle equation is given by:
$$x^2 + y^2 = 1$$
For an angle of \(45^{\circ}\) or \(\frac{\pi}{4}\) radians, the coordinates are:
$$x = \cos(\frac{\pi}{4})$$
$$y = \sin(\frac{\pi}{4})$$
Both \(\cos(\frac{\pi}{4})\) and \(\sin(\frac{\pi}{4})\) are equal to \(\frac{\sqrt{2}}{2}\).
Hence, the coordinates are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$.
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