Unit Circle

Find the tangent of the angle θ when θ is 45 degrees on the unit circle

To find the tangent of $45^\circ$ on the unit circle, we use the fact that $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

At $45^\circ$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$ and $\cos 45^\circ = \frac{\sqrt{2}}{2}$.

Therefore,

$$ \tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

Given a point on the unit circle where the secant of the angle is 3, find the angle in radians and degrees, and determine the corresponding coordinates on the unit circle

Given that $\sec \theta = 3$, we know that:

$$\sec \theta = \frac{1}{\cos \theta}$$

Solving for $\cos \theta$, we get:

$$\cos \theta = \frac{1}{3}$$

Using $\cos^{-1}(\frac{1}{3})$, we find:

$$\theta = \cos^{-1}(\frac{1}{3})$$

Converting to degrees:

$$\theta \approx 70.5288°$$

Since $\sec \theta = \sec (360° – \theta)$, the other solution is:

$$\theta = 360° – 70.5288° \approx 289.4712°$$

In radians, this is:

$$\theta \approx 1.23095 \text{ radians or } 5.05224 \text{ radians}$$

The corresponding coordinates on the unit circle are:

$$ (\cos (1.23095), \sin (1.23095)) = (\frac{1}{3}, \sqrt{1 – \frac{1}{9}}) = (\frac{1}{3}, \sqrt{\frac{8}{9}}) = (\frac{1}{3}, \frac{2\sqrt{2}}{3}) $$

and

$$ (\cos (5.05224), \sin (5.05224)) = (\frac{1}{3}, -\frac{2\sqrt{2}}{3}) $$

Find the sine and cosine of an angle given in radians on the unit circle

Given an angle \( \theta = \frac{\pi}{4} \), find the sine and cosine of the angle on the unit circle.

Using the unit circle, the coordinates of the point at \( \theta = \frac{\pi}{4} \) are given by: \( (\cos(\frac{\pi}{4}), \sin(\frac{\pi}{4})) \).

We know that:

$$ \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$

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

Therefore, the sine and cosine of the angle \( \theta = \frac{\pi}{4} \) are both \( \frac{\sqrt{2}}{2} \).

Find the values of sin(θ), cos(θ), and tan(θ) using the unit circle for θ = 135°

We start by locating the angle $\theta = 135°$ on the unit circle.

Since $135°$ is in the second quadrant, we use the reference angle $45°$ to find the values. The coordinates of the point on the unit circle at this angle are $\left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

Thus, $\sin(135°) = \frac{\sqrt{2}}{2}$, $\cos(135°) = -\frac{\sqrt{2}}{2}$, and $\tan(135°) = \frac{\sin(135°)}{\cos(135°)} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1$.

$$\sin(135°) = \frac{\sqrt{2}}{2}$$

$$\cos(135°) = -\frac{\sqrt{2}}{2}$$

$$\tan(135°) = -1$$

Find the value of cos(π/9) using the unit circle and trigonometric identities

To find the value of $\cos(\frac{\pi}{9})$, we can utilize the triple angle formula for cosine: $\cos(3\theta) = 4\cos^3(\theta) – 3\cos(\theta)$. Let $\theta = \frac{\pi}{9}$.

Therefore, $3\theta = \frac{3\pi}{9} = \frac{\pi}{3}$, and we know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$.

Substituting these values into the triple angle formula, we get:

$$\cos(\frac{\pi}{3}) = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9})$$

$$\frac{1}{2} = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9})$$

Let $x = \cos(\frac{\pi}{9})$, then we have the cubic equation:

$$\frac{1}{2} = 4x^3 – 3x$$

Rearranging gives:

$$4x^3 – 3x – \frac{1}{2} = 0$$

Using numerical methods, the solution is:

$$\cos(\frac{\pi}{9}) \approx 0.9848$$

Find the coordinates and trigonometric values of pi/10 on the unit circle

To find the exact coordinates of $\frac{\pi}{10}$ on the unit circle, we need to compute both the cosine and sine of this angle.

First, recall that the unit circle is defined by the equation $x^2 + y^2 = 1$ where $x = \cos(\theta)$ and $y = \sin(\theta)$. For the angle $\theta = \frac{\pi}{10}$, we have:

$$x = \cos\left(\frac{\pi}{10}\right)$$

$$y = \sin\left(\frac{\pi}{10}\right)$$

Using the half-angle and product-to-sum identities, we find:

$$\cos\left(\frac{\pi}{10}\right) = \sqrt{\frac{5 + \sqrt{5}}{8}}$$

$$\sin\left(\frac{\pi}{10}\right) = \sqrt{\frac{5 – \sqrt{5}}{8}}$$

Therefore, the coordinates of $\frac{\pi}{10}$ on the unit circle are:

$$\left(\sqrt{\frac{5 + \sqrt{5}}{8}}, \sqrt{\frac{5 – \sqrt{5}}{8}}\right)$$

How can you efficiently memorize the unit circle?

To efficiently memorize the unit circle, begin by understanding the key angles in radians and degrees. Break down the circle into quadrants, and focus on the primary angles: $0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{2}$, and $2\pi$. Draw connections between these angles and their sine and cosine values.

$$\text{For example, for } \frac{\pi}{6} (30^\circ), (\cos \frac{\pi}{6}, \sin \frac{\pi}{6}) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$$.

Visualize these values on the unit circle to create a mental map.

Find the sine and cosine of the angle \( \frac{5\pi}{4} \) on the unit circle

To find the sine and cosine of the angle \( \frac{5\pi}{4} \) on the unit circle, we start by locating the angle.

The angle \( \frac{5\pi}{4} \) is in the third quadrant of the unit circle.

We recognize that \( \frac{5\pi}{4} \) is the same as \( \pi + \frac{\pi}{4} \).

In the third quadrant, both sine and cosine are negative.

Now, take the reference angle \( \frac{\pi}{4} \), which has sine and cosine values of \( \frac{1}{\sqrt{2}} \) or \( \frac{\sqrt{2}}{2} \).

Since we are in the third quadrant, we apply the negative signs:

$$ \sin \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

$$ \cos \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

Find the coordinates on the unit circle for the angle θ = π/3

Given the angle $\theta = \pi/3$, we need to find the coordinates on the unit circle.

In the unit circle, the coordinates of an angle $\theta$ are $(\cos \theta, \sin \theta)$.

For $\theta = \pi/3$:

$$\cos(\pi/3) = \frac{1}{2}$$

$$\sin(\pi/3) = \frac{\sqrt{3}}{2}$$

Therefore, the coordinates are $(\frac{1}{2}, \frac{\sqrt{3}}{2})$.

Find the coordinates of a point on the flipped unit circle at a given angle

Given the angle $\theta = \frac{\pi}{3}$, find the coordinates of the corresponding point on the flipped unit circle where the x and y coordinates are switched.

The standard coordinates for $\theta = \frac{\pi}{3}$ on the unit circle are $(cos(\frac{\pi}{3}), sin(\frac{\pi}{3})) = (\frac{1}{2}, \frac{\sqrt{3}}{2})$.

For the flipped unit circle, the coordinates are switched, giving us $(y, x)$.

Therefore, the coordinates of the point at $\theta = \frac{\pi}{3}$ on the flipped unit circle are $$(\frac{\sqrt{3}}{2}, \frac{1}{2}).$$