Math

Find the value of tan(240) using the unit circle

To find the value of $\tan(240)$ using the unit circle, we first determine the corresponding point on the unit circle for an angle of 240 degrees.

240 degrees is in the third quadrant, where the tangent function is positive.

We can subtract 180 degrees to find the reference angle:

$$240^{\circ} – 180^{\circ} = 60^{\circ}$$

The reference angle is 60 degrees. The coordinates for 60 degrees on the unit circle are $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$.

Since we are in the third quadrant, both coordinates are negative:

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

The formula for tangent is:

$$ \tan(\theta) = \frac{y}{x} $$

Thus,

$$ \tan(240^{\circ}) = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} $$

Find the coordinates of a point on the unit circle that corresponds to an angle of 7π/4

To find the coordinates of the point on the unit circle that corresponds to an angle of $\frac{7\pi}{4}$, we use the sine and cosine functions:

$$x = \cos\left(\frac{7\pi}{4}\right)$$

$$y = \sin\left(\frac{7\pi}{4}\right)$$

Since $\frac{7\pi}{4}$ is in the fourth quadrant, we have:

$$\cos\left(\frac{7\pi}{4}\right) = \frac{1}{\sqrt{2}}$$

$$\sin\left(\frac{7\pi}{4}\right) = -\frac{1}{\sqrt{2}}$$

So, the coordinates are:

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

Find the derivative of cos(x^2) with respect to x

To find the derivative of $ \cos(x^2) $ with respect to $ x $, we use the chain rule:

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$$ \frac{d}{dx} \cos(u) = -\sin(u) \cdot \frac{du}{dx} $$

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Here, let $ u = x^2 $. Then:

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$$ \frac{du}{dx} = \frac{d}{dx}(x^2) = 2x $$

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Now apply the chain rule:

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$$ \frac{d}{dx} \cos(x^2) = -\sin(x^2) \cdot 2x $$

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The final derivative is:

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$$ \frac{d}{dx} \cos(x^2) = -2x \sin(x^2) $$

Determine the tangent values for the primary angles on the unit circle

To determine the tangent values for the primary angles on the unit circle, we need to evaluate the tangent function at $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ and $2\pi$.

$$ \text{tan}(0) = 0 $$

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

$$ \text{tan}\left(\frac{\pi}{4}\right) = 1 $$

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

$$ \text{tan}\left(\frac{\pi}{2}\right) = \text{undefined} $$

$$ \text{tan}(\pi) = 0 $$

$$ \text{tan}\left(\frac{3\pi}{2}\right) = \text{undefined} $$

$$ \text{tan}(2\pi) = 0 $$

How to find sine, cosine, and tangent for an angle using the unit circle?

To find the sine, cosine, and tangent of an angle using the unit circle, follow these steps:

1. Locate the angle on the unit circle.

2. Identify the coordinates $(x, y)$ of the point where the terminal side of the angle intersects the unit circle.

3. The coordinates correspond to $\cos(\theta)$ and $\sin(\theta)$ respectively:

$$ \cos(\theta) = x $$

$$ \sin(\theta) = y $$

4. Calculate $\tan(\theta)$ as follows:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{y}{x} $$

Determine the coordinates of the point where the angle θ = π/3 on the unit circle

First, recall that the unit circle has a radius of 1. For the angle $ \theta = \frac{\pi}{3} $, we use the definitions of sine and cosine:

$$ x = \cos(\theta) $$

$$ y = \sin(\theta) $$

When $ \theta = \frac{\pi}{3} $, we have:

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

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

Therefore, the coordinates of the point are:

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

Find the values of sin(θ) and cos(θ) where θ is 5π/4 radians on the unit circle

Given $\theta = \frac{5\pi}{4}$, we need to find the values of $\sin(\theta)$ and $\cos(\theta)$ on the unit circle.

The angle $\frac{5\pi}{4}$ is in the third quadrant where both sine and cosine are negative.

In the third quadrant, for an angle of $\frac{5\pi}{4}$,

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

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

Compute the integral of cos^2(t) on the unit circle

To compute the integral of $\cos^2(t)$ on the unit circle, we can use the double-angle identity for cosine:

$$\cos^2(t) = \frac{1 + \cos(2t)}{2}$$

Now, integrate:

$$\int_0^{2\pi} \cos^2(t) \, dt = \int_0^{2\pi} \frac{1 + \cos(2t)}{2} \, dt$$

Separate the integral:

$$= \frac{1}{2} \int_0^{2\pi} (1 + \cos(2t)) \, dt$$

Split it into two integrals:

$$= \frac{1}{2} \left( \int_0^{2\pi} 1 \, dt + \int_0^{2\pi} \cos(2t) \, dt \right)$$

The first integral is straightforward:

$$= \frac{1}{2} \left( 2\pi + 0 \right)$$

The second integral of cosine over a full period is zero:

$$= \pi$$

Find the coordinates of the point on the unit circle where the angle is 5π/6

To find the coordinates of the point on the unit circle where the angle is $\frac{5\pi}{6}$, we use the unit circle trigonometric identities for sine and cosine.

Since $\frac{5\pi}{6}$ is in the second quadrant:

The x-coordinate is:

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

The y-coordinate is:

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

So, the coordinates are:

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

Find the coordinates on the unit circle for an angle of pi/3

To find the coordinates on the unit circle for an angle of $ \frac{\pi}{3} $, we use the cosine and sine functions:

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

$$ y = \sin(\frac{\pi}{3}) $$

The values are:

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

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

Thus, the coordinates are:

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