Unit Circle

Determine the values of sin, cos, and tan for an angle of 7π/6 on the unit circle

First, locate the angle $\frac{7\pi}{6}$ on the unit circle. This angle corresponds to 210 degrees.

The coordinates of the point on the unit circle at this angle are:

$$ \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$

Thus:

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

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

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

Find the values of x for which cos(x) = -1/2 on the unit circle

To find the values of $\cos(x) = -\frac{1}{2}$ on the unit circle, we start by considering the unit circle where $\cos(\theta)$ is the x-coordinate of the point corresponding to the angle $\theta$:

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$$ \cos(x) = -\frac{1}{2} $$

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We know from trigonometric identities and the unit circle that:

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$$ \cos(120^\circ) = \cos\left( \frac{2\pi}{3} \right) = -\frac{1}{2} $$

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$$ \cos(240^\circ) = \cos\left( \frac{4\pi}{3} \right) = -\frac{1}{2} $$

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Therefore, the solutions in degrees are $120^\circ$ and $240^\circ$, and in radians they are:

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$$ x = \frac{2\pi}{3} + 2k\pi $$

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$$ x = \frac{4\pi}{3} + 2k\pi \quad \text{where } k \in \mathbb{Z} $$

Find the value of cos(theta) using the unit circle when theta = 5pi/4

To find the value of $ \cos(\theta) $ using the unit circle when $ \theta = \frac{5\pi}{4} $, we first locate this angle on the unit circle.

The angle $ \theta = \frac{5\pi}{4} $ lies in the third quadrant.

We know that $ \theta = \frac{5\pi}{4} $ is equivalent to $ 225^{\circ} $.

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

On the unit circle, the coordinates for $ 225^{\circ} $ are $ (-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}) $.

Therefore, $ \cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} $.

Find the equation of the line tangent to the unit circle at point (1, 0)

To find the equation of the tangent line to the unit circle at $(1, 0)$, we first recognize that the unit circle has the equation:

$$ x^2 + y^2 = 1 $$

The slope of the tangent line at any point $(x_0, y_0)$ on the circle can be found using implicit differentiation:

$$ 2x \x0crac{dx}{dx} + 2y \x0crac{dy}{dx} = 0 $$

At $(1, 0)$, this simplifies to:

$$ 2(1) + 2(0) \x0crac{dy}{dx} = 0 $$

Therefore,

$$ \x0crac{dy}{dx} = 0 $$

The slope is $0$, so the equation of the tangent line is:

$$ y = 0 $$

Find the value of sec(θ) when θ is on the unit circle

Given that $ \theta $ is an angle on the unit circle, we know that:

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

The cosine of $ \theta $ can be found using the coordinates (x, y) of the corresponding point on the unit circle, where x represents $ \cos(\theta) $.

Suppose $ \theta = \frac{\pi}{4} $, then:

$$ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $$

Therefore:

$$ \sec\left(\frac{\pi}{4}\right) = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} $$

Find the sine and cosine values for the angle 5π/4 using the unit circle

To find the sine and cosine values for the angle $ \frac{5\pi}{4} $ using the unit circle, first note that this angle is in the third quadrant.

In the unit circle, the reference angle for $ \frac{5\pi}{4} $ is $ \frac{\pi}{4} $.

The sine and cosine values for $ \frac{\pi}{4} $ are both $ \frac{\sqrt{2}}{2} $.

Since it is in the third quadrant, both sine and cosine are negative.

Thus, the sine and cosine values for $ \frac{5\pi}{4} $ are:

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

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

Find the coordinates of the point on the unit circle at 3π/4 radians

The unit circle has a radius of 1 and can be represented by the equation:

$$ x^2 + y^2 = 1 $$

At an angle of $\frac{3\pi}{4}$ radians, the coordinates can be determined using the sine and cosine functions:

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

$$ y = \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} $$

Therefore, the coordinates are:

$$ \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$

Evaluate the sine and cosine of \( \frac{\pi}{4} \)

To evaluate the sine and cosine of $ \frac{\pi}{4} $, we use the unit circle values:

The sine of $ \frac{\pi}{4} $ is:

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

The cosine of $ \frac{\pi}{4} $ is:

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

Determine the coordinates on the unit circle where the tangent line passes through the point (1,2)

To find the points on the unit circle where the tangent line passes through the point $(1, 2)$, we can use the following steps:

1. The equation of the unit circle is given by $$x^2 + y^2 = 1$$.

2. The slope of the tangent line at any point $(a, b)$ on the circle is $$-a/b$$.

3. The equation of the tangent line at $(a, b)$ can be written as:

$$y – b = -\frac{a}{b}(x – a)$$

4. Substitute the point $(1, 2)$ into this equation to find $(a, b)$:

$$2 – b = -\frac{a}{b}(1 – a)$$

5. Simplify and solve for $(a, b)$.

After solving, we get:

$$a = \frac{\sqrt{2}}{3}$$ and $$b = \frac{\sqrt{2}}{3}$$.

Thus, the coordinates are: $$(\frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3})$$ and $$(\frac{-\sqrt{2}}{3}, \frac{-\sqrt{2}}{3})$$.

What is the cosine of an angle in the unit circle if the sine is negative?

In the unit circle, if the $\sin(\theta)$ is negative, it means that the angle $\theta$ is in the third or fourth quadrant.

In both of these quadrants, the sine value is negative.

Cosine values in these quadrants can be positive (fourth quadrant) or negative (third quadrant).

Therefore, the $\cos(\theta)$ can be expressed as:

$$\cos(\theta) = \pm\sqrt{1 – \sin^2(\theta)}$$