Math

What is the value of cos(-π/3) using the unit circle?

To find the value of $\cos(-\frac{\pi}{3})$ using the unit circle, first recognize that the cosine function is an even function. This means that:

$$\cos(-x) = \cos(x)$$

Therefore:

$$\cos(-\frac{\pi}{3}) = \cos(\frac{\pi}{3})$$

From the unit circle, we know that:

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

Thus:

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

Find the coordinates of the point on the unit circle at which the angle is 7π/6

To find the coordinates of the point on the unit circle at which the angle is $ \frac{7\pi}{6} $, we use the following:

The unit circle has the equation:

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

The coordinates of a point on the unit circle are given by:

$$ (\cos(\theta), \sin(\theta)) $$

For $ \theta = \frac{7\pi}{6} $:

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

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

Therefore, the coordinates are:

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

Find the exact coordinates of the point(s) on the unit circle where the tangent line is vertical

The equation of the unit circle is given by:

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

We find the tangent line to be vertical when the derivative is undefined. Thus, we need to find the points where $ \x0crac{dy}{dx} $ is undefined.

Implicitly differentiate the unit circle equation with respect to $ x $:

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

Simplify and solve for $ \x0crac{dy}{dx} $:

$$ \x0crac{dy}{dx} = -\x0crac{x}{y} $$

The derivative is undefined when $ y = 0 $. Thus, we solve for $ x $:

When $ y = 0 $, substituting back into the original equation:

$$ x^2 = 1 $$

So, $ x = 1 $ or $ x = -1 $.

Therefore, the points are:

$(1,0)$ and $(-1,0)$.

Find the sine and cosine of the angle when the point is (1/2, √3/2) on the unit circle

The coordinates \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \) on the unit circle represent the cosine and sine of an angle:

$$\cos(\theta) = \frac{1}{2}$$ $$\sin(\theta) = \frac{\sqrt{3}}{2}$$

The angle in radians corresponding to these values is:

$$\theta = \frac{\pi}{3}$$

Evaluate the integral of sin(x) * cos(x) around the unit circle

To evaluate the integral of $ \sin(x) * \cos(x) $ around the unit circle, we can use the double-angle identity:

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

Now, we need to integrate from $ 0 $ to $ 2\pi $:

$$ \int_{0}^{2\pi} \sin(x) \cos(x) \, dx = \int_{0}^{2\pi} \frac{1}{2} \sin(2x) \, dx $$

Let $ u = 2x $, hence $ du = 2 \, dx $ and $ dx = \frac{1}{2} du $:

$$ \int_{0}^{2\pi} \frac{1}{2} \sin(2x) \, dx = \frac{1}{2} \int_{0}^{4\pi} \sin(u) \frac{1}{2} \, du $$

Combining constants:

$$ \frac{1}{4} \int_{0}^{4\pi} \sin(u) \, du $$

The integral of $ \sin(u) $ over one period is zero, and here we have two periods:

$$ \frac{1}{4} \left(0\right) = 0 $$

The integral evaluates to $ 0 $.

Find the terminal point on the unit circle for an angle of pi/6 radians

To find the terminal point on the unit circle for an angle of $ \frac{\pi}{6} $ radians, we use the unit circle definition:

The coordinates are given by $ ( \cos( \theta ), \sin( \theta ) ) $.

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

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

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

So, the terminal point is:

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

Identify the sine and cosine values for the angle π/4 on the unit circle

To find the sine and cosine values for the angle $ \frac{\pi}{4} $ on the unit circle, we use the definitions of sine and cosine:

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

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

Find the cosine of the angle at 3π/4 radians on the unit circle

The unit circle helps us find the cosine of an angle. For an angle of $ \frac{3π}{4} $ radians:

The reference angle is $ \x0crac{π}{4} $, and in the second quadrant, the cosine is negative.

So, $ \cos(\frac{3π}{4}) = -\cos(\frac{π}{4}) $

We know that $ \cos(\frac{π}{4}) = \frac{\sqrt{2}}{2} $

Therefore, $ \cos(\frac{3π}{4}) = -\frac{\sqrt{2}}{2} $

Prove that the equation $ x^2 + y^2 = 1 $ is satisfied by the coordinates of any point on the unit circle for a given angle \theta

To prove that the equation $ x^2 + y^2 = 1 $ is satisfied by the coordinates of any point on the unit circle for a given angle \theta , we start with the unit circle definition:

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On the unit circle, the coordinates of a point corresponding to an angle $ \theta $ are $ (\cos(\theta), \sin(\theta)) $.

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Consider the equation $ x^2 + y^2 = 1 $.

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Substitute $ x = \cos(\theta) $ and $ y = \sin(\theta) $:

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$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$

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This identity is known as the Pythagorean identity, and it holds true for all values of $ \theta $. Therefore, the equation $ x^2 + y^2 = 1 $ is satisfied by the coordinates of any point on the unit circle.

Evaluate the cosecant of an angle in the unit circle when its sine is equal to a rational value

To evaluate $ \csc(\theta) $ when $ \sin(\theta) $ is a rational value, let