Explore the unit circle and its relationship to angles, radians, trigonometric ratios, and coordinates in the coordinate plane.
Find the value of csc(θ) when θ = π/6 on the unit circle
To find the value of $csc(θ)$ when $θ = \frac{π}{6}$ on the unit circle, we first find the sine of $θ$:
$$ \sin(\frac{π}{6}) = \frac{1}{2} $$
Since $csc(θ) = \frac{1}{\sin(θ)}$, we have:
$$ csc(θ) = \frac{1}{\sin(\frac{π}{6})} = \frac{1}{\frac{1}{2}} = 2 $$
Find the exact value of cos(pi/9) using the unit circle and trigonometric identities
To find the exact value of $ \cos(\frac{\pi}{9}) $, we can use the triple-angle identity for cosine:
\n
$$ \cos(3\theta) = 4\cos^3(\theta) – 3\cos(\theta) $$
\n
Letting $ \theta = \frac{\pi}{9} $, we get:
\n
$$ \cos(\frac{\pi}{3}) = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9}) $$
\n
Since $ \cos(\frac{\pi}{3}) = \frac{1}{2} $, substituting in we have:
\n
$$ \frac{1}{2} = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9}) $$
\n
Let $ x = \cos(\frac{\pi}{9}) $, then the equation becomes:
\n
$$ \frac{1}{2} = 4x^3 – 3x $$
\n
Multiplying through by 2 to clear the fraction:
\n
$$ 1 = 8x^3 – 6x $$
\n
This is a cubic equation that can be solved for $ x = \cos(\frac{\pi}{9}) $ using numerical methods or by recognizing that:
\n
$$ \cos(\frac{\pi}{9}) = \frac{\sqrt{6} + \sqrt{2}}{4} $$
Evaluate the integral of sec(x) along the unit circle
To evaluate the integral of $ \sec(x) $ along the unit circle, we consider the parametrization of the unit circle. The unit circle can be parametrized as $ x = \cos(\theta) $ and $ y = \sin(\theta) $, where $ \theta $ ranges from $ 0 $ to $ 2\pi $.
The integral to evaluate becomes:
$$ \int_0^{2\pi} \sec(\cos(\theta)) \frac{d\theta}{d \theta} \ d\theta $$
We need to express $ \sec(\cos(\theta)) $ in terms of $ \theta $. However, since $ \sec(x) $ is not straightforward to integrate on the unit circle, it is more practical to use a different approach, often involving complex analysis or residue theorem.
Prove that sin(π/6) using the unit circle
To prove that $ \sin(\frac{\pi}{6}) $ using the unit circle, we start by locating the angle $ \frac{\pi}{6} $ on the unit circle.
The angle $ \frac{\pi}{6} $ corresponds to 30 degrees.
Using the unit circle, we see that the coordinates for this angle are $ (\frac{\sqrt{3}}{2}, \frac{1}{2}) $.
Since the sine of an angle is the y-coordinate on the unit circle, we have:
$$ \sin(\frac{\pi}{6}) = \frac{1}{2} $$
Determine the values of cos(θ) and sin(θ) using the unit circle when 0 ≤ θ ≤ 2π and θ is a solution to the equation tan(θ) = √3
The equation $\tan(\theta) = \sqrt{3}$ implies that:
$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \sqrt{3}$$
This happens at $\theta = \frac{\pi}{3}$ and $\theta = \frac{4\pi}{3}$ within the interval $0 ≤ \theta ≤ 2\pi$.
At $\theta = \frac{\pi}{3}$:
$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$
At $\theta = \frac{4\pi}{3}$:
$$\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}, \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$
Thus, the values are:
$$\theta = \frac{\pi}{3}: \cos(\theta) = \frac{1}{2}, \sin(\theta) = \frac{\sqrt{3}}{2}$$
$$\theta = \frac{4\pi}{3}: \cos(\theta) = -\frac{1}{2}, \sin(\theta) = -\frac{\sqrt{3}}{2}$$
Find the value of angle θ where cos(θ) = -1/2 on the unit circle
The cosine function represents the x-coordinate on the unit circle. Thus, finding $ \cos(\theta) = -\frac{1}{2} $ involves finding the angles where the x-coordinate is -1/2. On the unit circle, this occurs at:
$$ \theta = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{4\pi}{3} + 2k\pi $$
for any integer $ k $.
Find the angle in radians and degrees for the point (-1/2, -√3/2) on the unit circle
We need to find the angle corresponding to the point $ \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) $ on the unit circle. This point lies in the third quadrant where both sine and cosine are negative. The reference angle is given by:
$$ \text{Reference angle} = \arccos\left( \frac{1}{2} \right) = \frac{\pi}{3} $$
Since the point is in the third quadrant, the angle in radians is:
$$ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $$
To convert this to degrees:
$$ \theta = \frac{4\pi}{3} \times \frac{180}{\pi} = 240^{\circ} $$
Hence, the angle is $ \frac{4\pi}{3} $ radians or $ 240^{\circ} $.
Find the value of arcsin(1/2) in radians using the unit circle
To find the value of $ \arcsin(\frac{1}{2}) $, consider the unit circle and the definition of arcsin. The arcsin function outputs the angle whose sine is the given value within the range $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $.
For $ \arcsin(\frac{1}{2}) $, we need to find the angle $ \theta $ such that $ \sin(\theta) = \frac{1}{2} $. On the unit circle, $ \sin(30^{\circ}) = \frac{1}{2} $ or equivalently, in radians:
$$ \theta = \frac{\pi}{6} $$
Thus, the value of $ \arcsin(\frac{1}{2}) $ is:
$$ \arcsin(\frac{1}{2}) = \frac{\pi}{6} $$
Find the value of cosine at $\theta = \frac{3\pi}{4}$
The unit circle helps us locate the angle $\theta = \frac{3\pi}{4}$ which lies in the second quadrant. The reference angle for $\theta = \frac{3\pi}{4}$ is:
$$\pi – \frac{3\pi}{4} = \frac{\pi}{4}$$
In the second quadrant, the cosine of an angle is negative:
$$\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right)$$
Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we have:
$$\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$
Find the coordinates of the point where the angle π/4 intersects the unit circle
To find the coordinates of the point where the angle $ \frac{\pi}{4} $ intersects the unit circle, we use the unit circle definition. The unit circle has a radius of 1, and the coordinates of the points on the circle are given by the cosine and sine of the angle.
For the angle $ \frac{\pi}{4} $, we have:
$$ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
$$ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
Therefore, the coordinates are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
Start Using PopAi Today
Suggested Content
More >
Answer 1 To find the value of $\tan(\theta)$ for $\theta$ on the unit circle, consider the point $P(x, y)$ on the circle corresponding to the angle $\theta$. The tangent of $\theta$ is given by the ratio of the y-coordinate to the x-coordinate, i.e.,...
Answer 1 Given the point \((x, y)\) on a unit circle, determine the quadrant in which the point lies. The unit circle has a radius of 1. The quadrants are defined as follows: - Quadrant I: \((x > 0, y > 0)\) - Quadrant II: \((x < 0, y > 0)\) -...
Answer 1 For $\theta = \frac{\pi}{4}$:On the unit circle, at $\theta = \frac{\pi}{4}$, the coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.The tangent function is given by $\tan(\theta) = \frac{y}{x}$.Thus, $ \tan(\frac{\pi}{4}) =...
Answer 1 To convert degrees to radians, we use the conversion factor \( \frac{\pi}{180} \).$150^{\circ} \times \frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6}$The coordinates on the unit circle for \( \theta = \frac{5\pi}{6} \) are given by...
Answer 1 To solve this problem, we need to determine the angle $\theta$ on the unit circle. Given: $\cos(\theta) = \frac{1}{2}$ $\sin(\theta) = \frac{\sqrt{3}}{2}$ On the unit circle, these values correspond to the angle $\theta = 60^{\circ}$ or...
Answer 1 To find the sine and cosine values at different points on the unit circle, we can use the angle in radians.1. For the angle $\frac{\pi}{6}$ radians: The coordinates on the unit circle are $\left( \cos \frac{\pi}{6}, \sin \frac{\pi}{6}...