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Find the angle in degrees corresponding to 7π/6 radians on the unit circle
To convert $\frac{7\pi}{6}$ radians to degrees, we use the conversion factor:
$$ 180^{\circ} = \pi \text{ radians} $$
Thus,
$$ \frac{7\pi}{6} \times \frac{180^{\circ}}{\pi} = 210^{\circ} $$
The angle in degrees is:
$$ 210^{\circ} $$
Find the coordinates on the unit circle where the tangent of the angle is 1
To find the coordinates on the unit circle where $ \tan(\theta) = 1 $, we need to determine the angles $\theta $ for which this condition holds. We know that:
$$ \tan(\theta) = \frac {\sin(\theta)}{\cos(\theta)} $$
For the tangent to be 1, the sine and cosine must be equal. This occurs at angles:
$$ \theta = \frac {\pi}{4} \text{ and } \theta = \frac {5\pi}{4} $$
Now, we find the coordinates on the unit circle for these angles:
$$ \text{At } \theta = \frac {\pi}{4}, \text{ the coordinates are } \left( \frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2} \right) $$
$$ \text{At } \theta = \frac {5\pi}{4}, \text{ the coordinates are } \left( -\frac {\sqrt {2}}{2}, -\frac {\sqrt {2}}{2} \right) $$
Thus, the coordinates on the unit circle where $ \tan(\theta) = 1 $ are:
$$ \left( \frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2} \right) $$
and
$$ \left( -\frac {\sqrt {2}}{2}, -\frac {\sqrt {2}}{2} \right) $$
Find the sine value at π/3 on the unit circle
To find the sine value at $ \frac{\pi}{3} $ on the unit circle, we use the unit circle definition. The angle $ \frac{\pi}{3} $ is equivalent to 60 degrees.
On the unit circle, the sine value at 60 degrees is:
$$ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $$
Find the sine and cosine of the angle θ on the unit circle when θ = 5π/4
To find the sine and cosine of the angle $\theta = \frac{5\pi}{4}$ on the unit circle, we use the definitions of the trigonometric functions on the unit circle. The angle $\frac{5\pi}{4}$ is in the third quadrant.
For angles in the third quadrant, both sine and cosine are negative. The reference angle for $\theta = \frac{5\pi}{4}$ is $\frac{\pi}{4}$.
The sine and cosine of $\frac{\pi}{4}$ are both $\frac{\sqrt{2}}{2}$.
Thus:
$$\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$
$$\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$
Find the coordinates of a point on the unit circle that corresponds to an angle of π/3 radians
To find the coordinates of a point on the unit circle corresponding to an angle of $ \frac{\pi}{3} $ radians:
We can use the unit circle definitions for sine and cosine.
$$ x = \cos \left( \frac{\pi}{3} \right) $$
$$ y = \sin \left( \frac{\pi}{3} \right) $$
Since $ \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} $ and $ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $, the coordinates are:
$$ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $$
Find the angle θ in radians where cos(θ) = -1/2 and 0 ≤ θ < 2π
To find the angle $ \theta $ in radians where $ \cos(\theta) = -\frac{1}{2} $ and $ 0 \leq \theta < 2\pi $, we look for the points on the unit circle where the x-coordinate is -1/2.
These points correspond to angles in the second and third quadrants.
In the second quadrant, the angle is:
$$ \theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $$
In the third quadrant, the angle is:
$$ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $$
Therefore, the angles are:
$$ \theta = \frac{2\pi}{3} \text{ and } \frac{4\pi}{3} $$
Find the general solution of the equation tan(x) = cot(2x) on the unit circle
To solve the equation $ \tan(x) = \cot(2x) $ on the unit circle, we start by expressing $ \cot(2x) $ in terms of $ \tan(2x) $:
$$ \cot(2x) = \frac{1}{\tan(2x)} $$
The equation becomes:
$$ \tan(x) = \frac{1}{\tan(2x)} $$
Using the double-angle identity for tangent:
$$ \tan(2x) = \frac{2 \tan(x)}{1 – \tan^2(x)} $$
Substitute this back into the equation:
$$ \tan(x) = \frac{1}{\frac{2 \tan(x)}{1 – \tan^2(x)}} $$
Simplify the equation:
$$ \tan(x) = \frac{1 – \tan^2(x)}{2 \tan(x)} $$
Rearrange the equation:
$$ 2 \tan^2(x) = 1 – \tan^2(x) $$
Combine like terms:
$$ 3 \tan^2(x) = 1 $$
Solve for $ \tan(x) $:
$$ \tan(x) = \pm \frac{1}{\sqrt{3}} $$
Therefore, the general solution on the unit circle is:
$$ x = n\pi + (-1)^n \frac{\pi}{6} $$ where $ n $ is an integer.
Find the coordinates of the point on the unit circle corresponding to an angle of \( \frac{\pi}{4} \)
To find the coordinates of the point on the unit circle corresponding to an angle of $ \frac{\pi}{4} $, we use the unit circle definition:
The unit circle equation is: $$ x^2 + y^2 = 1 $$
At an angle of $ \frac{\pi}{4} $, both cosine and sine values are equal. Hence:
$$ x = \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$
$$ y = \sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$
So, the coordinates are: $$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
Determine the coordinates of a point on the unit circle at an angle of 11π/6
To determine the coordinates of a point on the unit circle at an angle of $$ \frac{11\pi}{6} $$, we use the unit circle properties.
The coordinates can be found using the cosine and sine of the angle:
$$ x = \cos \left( \frac{11\pi}{6} \right) $$
$$ y = \sin \left( \frac{11\pi}{6} \right) $$
Since $$ \frac{11\pi}{6} $$ is in the fourth quadrant, we know:
$$ \cos \left( \frac{11\pi}{6} \right) = \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} $$
$$ \sin \left( \frac{11\pi}{6} \right) = -\sin \left( \frac{\pi}{6} \right) = -\frac{1}{2} $$
So the coordinates are:
$$ \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$
Solve for the exact values of all angles θ in the interval [0, 2π) that satisfy cos(θ) = -1/2
To find the exact values of all angles $ \theta $ in the interval $ [0, 2\pi) $ that satisfy $ \cos(\theta) = -\frac{1}{2} $, we use the unit circle. The cosine value of $ -\frac{1}{2} $ corresponds to angles in the second and third quadrants. The reference angle is $ \frac{\pi}{3} $.
The angles are:
Thus, the solutions are:
$$ \theta = \frac{2\pi}{3}, \frac{4\pi}{3} $$
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