Unit Circle

Find the exact values of the sine and cosine of an angle using the unit circle

To find the exact values of $\sin(\theta)$ and $\cos(\theta)$ using the unit circle, let’s consider $\theta = \frac{5\pi}{6}$.

First, we know that $\frac{5\pi}{6}$ is in the second quadrant.

In the second quadrant, sine is positive and cosine is negative.

Using the reference angle $\frac{\pi}{6}$, we have:

$$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi – \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$

and

$$\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\pi – \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$

Find the value of cos θ on the unit circle in the complex plane when θ = π/3

To find the value of $\cos \theta$ on the unit circle, we use the unit circle definition where the coordinates are $(\cos \theta, \sin \theta)$.

For $\theta = \pi/3$, the coordinates on the unit circle are:

$$\left( \cos \frac{\pi}{3}, \sin \frac{\pi}{3} \right) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$$

Therefore,

$$\cos \frac{\pi}{3} = \frac{1}{2}$$

Find all angles θ in the interval [0, 2π) where cos(θ + π/6) = √3/2

To solve for all angles \( \theta \) in the interval \([0, 2\pi)\) where \( \cos(\theta + \pi/6) = \sqrt{3}/2 \), we first identify the standard angles where cosine equals \( \sqrt{3}/2 \). These angles are:

$$\alpha = 0 \text{ or } \alpha = 2\pi$$

Next, we set up the equation:

$$\theta + \pi/6 = 0 + 2k\pi \text{ or } \theta + \pi/6 = 2\pi + 2k\pi$$

where \( k \) is an integer.

Solving these equations for \( \theta \):

\( \theta = -\pi/6 + 2k\pi \) or \( \theta = 11\pi/6 + 2k\pi \)

Since \( \theta \) must be in the interval \([0, 2\pi)\), we find specific solutions by setting \( k = 0 \):

\( \theta = -\pi/6 = 11\pi/6 \) (not in the interval)

and \( \theta = 11\pi/6 \text{ (valid)} $$

Find the exact values of sine, cosine, and tangent for the angle θ where θ is in the third quadrant, and the terminal side of θ passes through the point (-3, -4) on the unit circle

Given point $(-3, -4)$, we first calculate the radius r:

$$ r = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

In the unit circle, the radius (or hypotenuse) is 1. So, we need to normalize the coordinates to fit the unit circle.

$$ x = \frac{-3}{5} $$

$$ y = \frac{-4}{5} $$

Thus, the coordinates on the unit circle are $(-\frac{3}{5}, -\frac{4}{5})$.

Therefore,

$$ \sin(\theta) = -\frac{4}{5} $$

$$ \cos(\theta) = -\frac{3}{5} $$

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} $$

Calculate the trigonometric values at specific angles on the unit circle

Given the angle $ \theta = \frac{3\pi}{4} $, find the values of $ \sin \theta $, $ \cos \theta $, and $ \tan \theta $.

Step 1: Identify the reference angle. The reference angle for $ \theta = \frac{3\pi}{4} $ is $ \frac{\pi}{4} $.

Step 2: Determine the sine, cosine, and tangent values of the reference angle. For $ \theta = \frac{\pi}{4} $, $ \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} $, $ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $, and $ \tan \frac{\pi}{4} = 1 $.

Step 3: Determine the signs of these values in the second quadrant. In the second quadrant, sine is positive, cosine is negative, and tangent is negative.

Therefore, $ \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} $, $ \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} $, and $ \tan \frac{3\pi}{4} = -1 $.

For a unit circle centered at the origin, consider a point P on the circle described by the angle θ measured from the positive x-axis in the counterclockwise direction Determine the coordinates of the point Q, which is the reflection of P across the line

The coordinates of the point $P$ on the unit circle can be expressed as $P(\cos\theta, \sin\theta)$. To find the coordinates of the reflection of $P$ across the line $y=x$, we interchange the x and y coordinates of $P$. Therefore, the coordinates of $Q$ are $Q(\sin\theta, \cos\theta)$.
Let’s verify:
Given $P(\cos\theta, \sin\theta)$, reflecting across $y=x$ gives $Q(\sin\theta, \cos\theta)$.
Thus, the coordinates of $Q$ are $Q(\sin\theta, \cos\theta)$.

Finding Trigonometric Values at Specific Angles on the Unit Circle

To find the trigonometric values of an angle of 150 degrees on the unit circle, we can use the angle relationship and symmetry of the unit circle.

1. Convert the angle to radians: $150° = \frac{5\pi}{6}$ radians.

2. Recall that on the unit circle, the coordinates of a point corresponding to an angle θ in radians are $(\cos θ, \sin θ)$.

3. Since $\frac{5\pi}{6}$ is in the second quadrant, where the cosine value is negative and the sine value is positive, we can use the reference angle of $\pi – \frac{5\pi}{6} = \frac{\pi}{6}$.

4. Therefore, $\cos \frac{5\pi}{6} = -\cos \frac{\pi}{6} = – \frac{\sqrt{3}}{2}$ and $\sin \frac{5\pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2}$.

Hence, the coordinates for 150 degrees on the unit circle are $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.

Find the secant of an angle on the unit circle for θ = 45 degrees

To find the secant of the angle $\theta = 45^{\circ}$ on the unit circle, we use the relationship between secant and cosine.

The secant of an angle is the reciprocal of its cosine:

$\sec(\theta) = \frac{1}{\cos(\theta)}$.

For $\theta = 45^{\circ}$, the cosine value is $\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$.

Therefore,

$$\sec(45^{\circ}) = \frac{1}{\cos(45^{\circ})} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$$.

So, the secant of $45^{\circ}$ is $\sqrt{2}$.

Find the Values of tan(θ) Given a Point on the Unit Circle

Given a point $P(a, b)$ on the unit circle, find the value of $\tan(\theta)$ where $\theta$ is the angle formed by the radius to the point P and the positive x-axis.

Solution: We know that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. Since P(a, b) lies on the unit circle, $\cos(\theta) = a$ and $\sin(\theta) = b$. Therefore, $\tan(\theta) = \frac{b}{a}$.

So, the value of $\tan(\theta)$ is $\frac{b}{a}$.

Find the sine, cosine, and tangent values for the angle $\frac{\pi}{6}$ on the unit circle

To solve this, we need to find the sine, cosine, and tangent values for the angle $\frac{\pi}{6}$ on the unit circle.

The angle $\frac{\pi}{6}$ corresponds to 30 degrees.

Using the unit circle, we know that:

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

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

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

So, the values are:

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

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

$$ \tan \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{3} $$