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Find the coordinates on the unit circle for the angles
Find the coordinates on the unit circle for the angle $ \theta = \frac{\pi}{4} $:
The coordinates are given by $ (\cos(\theta), \sin(\theta)) $.
For $ \theta = \frac{\pi}{4} $:
$$ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
$$ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
The coordinates are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.
Find the tangent values at 0, π/4, and π/3 on the unit circle
To find the tangent values at points $0$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$ on the unit circle, we use the tangent function $tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$:
1. At $\theta = 0$:
$$ \tan(0) = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0 $$
2. At $\theta = \frac{\pi}{4}$:
$$ \tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$
3. At $\theta = \frac{\pi}{3}$:
$$ \tan\left(\frac{\pi}{3}\right) = \frac{\sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} $$
Find the coordinates where $ \cos(\theta) = \sin(\theta) $ on the unit circle
To find the coordinates where $ \cos(\theta) = \sin(\theta) $ on the unit circle, we start from the equation:
$$ \cos(\theta) = \sin(\theta) $$
Since both cosine and sine are equal, we can express this as:
$$ \cos(\theta) = \sin(\theta) $$
Divide both sides by $ \cos(\theta) $:
$$1 = \tan(\theta) $$
This implies
$$ \theta = \frac{\pi}{4} + n\pi $$
for integer values of n. The corresponding coordinates on the unit circle are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$ and $$ \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $$
Calculate the exact value of sin(7π/6) using the unit circle
To determine the exact value of $\sin\left(\frac{7\pi}{6}\right)$ using the unit circle, first note that $\frac{7\pi}{6}$ is in the third quadrant.
In the third quadrant, the sine function is negative.
Now, find the reference angle for $\frac{7\pi}{6}$:
$$ 7\pi / 6 – \pi = \pi / 6 $$
The reference angle is $\pi / 6$, whose sine value is $\frac{1}{2}$.
Since sine is negative in the third quadrant:
$$ \sin \left( \frac{7\pi}{6} \right) = -\frac{1}{2} $$
Find the value of arcsin(sqrt(3)/2) based on the unit circle
To find the value of $ \arcsin(\sqrt{3}/2) $, we need to locate where $ \sin(x) = \sqrt{3}/2 $ on the unit circle.
On the unit circle, $ \sin(x) = \sqrt{3}/2 $ at the angles:
$$ x = \frac{\pi}{3} $$
and
$$ x = \frac{2\pi}{3} $$
So,
$$ \arcsin(\sqrt{3}/2) = \frac{\pi}{3} $$
in the primary range of arcsin, which is $ [ -\frac{\pi}{2}, \frac{\pi}{2} ] $.
Find the value of arccos(-1/2) and verify it using the unit circle
To find the value of $\arccos(-1/2)$, we need to determine the angle in the unit circle whose cosine is $-1/2$.
\n
From the unit circle, we know that:
\n
$$ \cos(\pi – \frac{\pi}{3}) = \cos(\frac{2\pi}{3}) = -1/2 $$
\n
Hence, the value of $\arccos(-1/2)$ is $\frac{2\pi}{3}$.
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Verification:
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Consider the angle $\frac{2\pi}{3}$ in the unit circle, its cosine value is:
\n
$$ \cos(\frac{2\pi}{3}) = -1/2 $$
\n
This matches our original value, verifying that $\arccos(-1/2) = \frac{2\pi}{3}$.
Determine the quadrant of a point on the unit circle given by an angle
Given an angle $ \theta $, we need to determine in which quadrant the corresponding point on the unit circle lies. The quadrants are determined as follows:
1. If $ 0 \leq \theta < \frac{\pi}{2} $, the point is in the first quadrant.
2. If $ \frac{\pi}{2} \leq \theta < \pi $, the point is in the second quadrant.
3. If $ \pi \leq \theta < \frac{3\pi}{2} $, the point is in the third quadrant.
4. If $ \frac{3\pi}{2} \leq \theta < 2\pi $, the point is in the fourth quadrant.
Determine the cotangent of an angle on the unit circle
The cotangent of an angle $ \theta $ on the unit circle is given by:
$$ \cot( \theta ) = \frac{1}{\tan( \theta )} = \frac{\cos( \theta )}{\sin( \theta )} $$
Let
Determine the equation of a unit circle and explain the geometric significance
The equation of a unit circle centered at the origin is given by:
$$ x^2 + y^2 = 1 $$
This equation signifies that any point $ (x, y) $ on the unit circle is at a distance of 1 unit from the origin. The radius of the circle is always 1.
Find the value of tan(θ) using the unit circle when θ = 3π/4
We need to find the value of $ \tan(\theta) $ where $ \theta = \frac{3\pi}{4} $ using the unit circle. The coordinates of the point on the unit circle corresponding to $ \theta = \frac{3\pi}{4} $ are:
$$ \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
Recall that $ \tan(\theta) = \frac{y}{x} $. Therefore:
$$ \tan \left( \frac{3\pi}{4} \right) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$
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