Unit Circle

Explore the unit circle and its relationship to angles, radians, trigonometric ratios, and coordinates in the coordinate plane.

Find the coordinates where $ \cos(\theta) = \sin(\theta) $ on the unit circle

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

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

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

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 $\frac{2\pi}{3}$.

\n

Verification:

\n

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

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

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

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

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:

$$ \tan \left( \frac{3\pi}{4} \right) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

Determine the trigonometric identity of sin(θ) using the unit circle

Determine the trigonometric identity of sin(θ) using the unit circle

To determine the trigonometric identity of $ \sin(\theta) $ using the unit circle, we start by understanding the unit circle definition:

The unit circle is a circle with a radius of $1$ centered at the origin $(0, 0)$.

For any angle $\theta$ measured from the positive x-axis, the coordinates of the point where the terminal side of $\theta$ intersects the unit circle are given by $(\cos(\theta), \sin(\theta))$.

Therefore, the identity for $\sin(\theta)$ is the y-coordinate of this intersection point:

$$ \sin(\theta) = y $$

Where $y$ is the y-coordinate of the intersection point.

To provide a concrete example, if $\theta = \frac{\pi}{4}$, the coordinates of the intersection point are $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, so:

$$ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$

Find the values of sin(θ), cos(θ), and tan(θ) for θ = 7π/6 using the unit circle

Find the values of sin(θ), cos(θ), and tan(θ) for θ = 7π/6 using the unit circle

To find the values of $ \sin(\theta) $, $ \cos(\theta) $, and $ \tan(\theta) $ for $ \theta = \frac{7\pi}{6} $ using the unit circle, we start by locating the angle on the unit circle:

$ \theta = \frac{7\pi}{6} $ corresponds to an angle in the third quadrant, where both sine and cosine values are negative.

In the unit circle, for $ \theta = \frac{7\pi}{6} $:

$$ \sin\left( \frac{7\pi}{6} \right) = -\frac{1}{2} $$

$$ \cos\left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2} $$

To find the tangent, use: $$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$

$$ \tan\left( \frac{7\pi}{6} \right) = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

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