Homework

Find the point on the unit circle where the sine value is negative and the cosine value is positive

The unit circle is defined as the set of all points $(x, y)$ such that:

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

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In the unit circle, the sine value corresponds to the y-coordinate and the cosine value corresponds to the x-coordinate. We need to find a point where:

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$$ y < 0 $$

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$$ x > 0 $$

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One such point is:

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

Determine the coordinates of points on the unit circle that satisfy the given equation

To determine the coordinates of points on the unit circle that satisfy the equation $ \cos^2(\theta) – \sin^2(\theta) = 0 $:

First, we recall the Pythagorean identity: $$ \cos^2(\theta) + \sin^2(\theta) = 1 $$

Given the equation: $$ \cos^2(\theta) – \sin^2(\theta) = 0 $$

This can be rewritten as: $$ \cos^2(\theta) = \sin^2(\theta) $$

Taking the square root of both sides gives: $$ \cos(\theta) = \pm \sin(\theta) $$

We consider the positive and negative cases separately.

For $ \cos(\theta) = \sin(\theta) $:

$$ \theta = \frac{\pi}{4} + k \pi $$

Where $ k $ is any integer.

For $ \cos(\theta) = -\sin(\theta) $:

$$ \theta = \frac{3\pi}{4} + k \pi $$

Where $ k $ is any integer.

Thus, the coordinates of the points on the unit circle are:

$$ ( \cos(\frac{\pi}{4} + k \pi), \sin(\frac{\pi}{4} + k \pi) ) $$

$$ ( \cos(\frac{3\pi}{4} + k \pi), \sin(\frac{3\pi}{4} + k \pi) ) $$

What is the value of cos(π/4) on the unit circle?

Find the coordinates of the point on the unit circle where the inverse sine function is equal to 1/2

To find the coordinates of the point on the unit circle where the inverse sine function, $ \sin^{-1}(x) $, is equal to $ \frac{1}{2} $:

We need to solve the equation:

$$ \sin^{-1}(y) = \frac{1}{2} $$

The angle whose sine is $ \frac{1}{2} $ is:

$$ \theta = \frac{\pi}{6} $$

On the unit circle, the coordinates corresponding to this angle are:

$$ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) $$

Evaluating the trigonometric functions, we get:

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

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

So, the coordinates are:

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

Determine the value of sec(θ) given cos(θ) = -1/sqrt(2) and θ is in the third quadrant

Given that $\cos(\theta) = -\frac{1}{\sqrt{2}}$ and $\theta$ is in the third quadrant, we start by using the identity:

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

Substituting the given value:

$$\sec(\theta) = \frac{1}{-\frac{1}{\sqrt{2}}}$$

We simplify the fraction:

$$\sec(\theta) = -\sqrt{2}$$

Thus, the value of $\sec(\theta)$ is $-\sqrt{2}$.

Find the sine and cosine values at t = pi/4 on the unit circle

To find the sine and cosine values at $ t = \frac{\pi}{4} $ on the unit circle, we use the following values:

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

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

Thus, the sine and cosine values are:

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

Identify the coordinates on the unit circle for θ = π/4

To find the coordinates on the unit circle for $ \theta = \frac{\pi}{4} $, we use the unit circle properties.

For $ \theta = \frac{\pi}{4} $, the coordinates are given by:

$$ (\cos(\frac{\pi}{4}), \sin(\frac{\pi}{4})) $$

From trigonometric values, we know:

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

So the coordinates are:

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

Find the values of sin and cos at the given angle on the unit circle

To find the values of $\sin$ and $\cos$ at an angle of $\theta = \frac{\pi}{3}$ on the unit circle:

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

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

Find the angle in radians for a point on the unit circle where the x-coordinate is 1/2

To find the angle $\theta$ in radians for a point on the unit circle where the $x$-coordinate is $\frac{1}{2}$, we consider the cosine function:

$$\cos(\theta) = \frac{1}{2}$$

The angles that satisfy this equation are:

$$\theta = \frac{\pi}{3}$$ and $$\theta = -\frac{\pi}{3}$$ or equivalently $$\theta = \frac{5\pi}{3}$$