Homework

Find the values of tan(x) at different positions on the unit circle

To find the values of $ \tan(x) $ at different positions on the unit circle, we use the definition:

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

First, let

Calculate the integral of 1/(x + sqrt(x^2 – 1)) over the interval [-1, 1]

To calculate the integral:

$$ \int_{-1}^{1} \frac{1}{x + \sqrt{x^2 – 1}} dx $$

First, consider the substitution $ x = \cosh(t) $, which implies $ dx = \sinh(t) dt $.

When $ x = -1 $, $ t = i \pi $ and when $ x = 1 $, $ t = 0 $:

$$ \int_{i \pi}^{0} \frac{1}{\cosh(t) + \sinh(t)} \sinh(t) dt $$

Knowing that $ \cosh(t) + \sinh(t) = e^t $, the integral becomes:

$$ \int_{i \pi}^{0} \frac{\sinh(t)}{e^t} dt = \int_{i \pi}^{0} e^{-t} dt $$

Evaluating this gives:

$$ [ -e^{-t} ]_{i \pi}^{0} = -e^{0} + e^{-i \pi} = -1 + (-1) = -2 $$

Find the minimum value of cos(θ1+θ2+θ3) where θ1, θ2, θ3 are angles on the unit circle satisfying specific conditions

Let

Find the sine of a negative angle on the unit circle

On the unit circle, the sine of a negative angle $ \theta $ is given by:

$$ \sin(-\theta) = -\sin(\theta) $$

For example, if $ \theta = 30^{\circ} $, then:

$$ \sin(-30^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2} $$

Find the values of sin(θ), cos(θ), and tan(θ) at θ = π/4 on the unit circle

To find the values of $ \sin(\theta) $, $ \cos(\theta) $, and $ \tan(\theta) $ at $ \theta = \frac{\pi}{4} $ on the unit circle, we use the standard trigonometric values:

\n

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

\n

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

\n

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

Determine the sine and cosine of an angle in the unit circle in the second quadrant

An angle $ \theta $ in the second quadrant of the unit circle ranges from $ 90^\circ $ to $ 180^\circ $ (or $ \frac{\pi}{2} $ to $ \pi $ radians). In this range, the sine of the angle is positive, and the cosine is negative.

For example, for $ \theta = 120^\circ $ (or $ \frac{2\pi}{3} $ radians):

$$ \sin(120^\circ) = \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} $$

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

Thus, the sine and cosine of an angle $ \theta $ in the second quadrant are:

$$ \sin(\theta) > 0 $$

$$ \cos(\theta) < 0 $$

Determine the coordinates and angles for points on the unit circle where the cosine value is 1/2

To determine the coordinates where $\cos(\theta) = \frac{1}{2}$ on the unit circle, we need to find $\theta$ such that:

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

The angles that satisfy this condition are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$. The corresponding coordinates are:

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

Find the angle in radians where the coordinates on the unit circle are (sqrt(3)/2, 1/2)

The coordinates $ \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) $ on the unit circle correspond to the angle $ \frac{\pi}{6} $ radians. We can confirm this by noting that $ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} $ and $ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} $.

Find the value of the integral of cot(x) from 0 to pi/4 using the unit circle

To find the value of the integral of $ \cot(x) $ from $ 0 $ to $ \frac{\pi}{4} $ using the unit circle, we first express cotangent in terms of sine and cosine:

$$ \cot(x) = \frac{\cos(x)}{\sin(x)} $$

The integral becomes:

$$ \int_{0}^{\frac{\pi}{4}} \cot(x) \, dx = \int_{0}^{\frac{\pi}{4}} \frac{\cos(x)}{\sin(x)} \, dx $$

Let $ u = \sin(x) $. Then $ du = \cos(x) \, dx $.

Now, change the limits of integration accordingly: when $ x = 0 $, $ u = \sin(0) = 0 $, and when $ x = \frac{\pi}{4} $, $ u = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $.

Thus, the integral becomes:

$$ \int_{0}^{\frac{\sqrt{2}}{2}} \frac{1}{u} \, du = \left. \ln|u| \right|_{0}^{\frac{\sqrt{2}}{2}} $$

Evaluating this, we get:

$$ \ln \left( \frac{\sqrt{2}}{2} \right) – \ln(0) $$

Note that $ \ln(0) $ is undefined, suggesting an improper integral. Thus, we interpret the limit at $ u \to 0^{+} $:

$$ \lim_{u \to 0^{+}} \ln(u) = -\infty $$

The final value of the integral is:

$$ \boxed{-\infty} $$

Find the value of sec(π/4)

To find the value of $ \sec(\frac{\pi}{4}) $, we first find the value of $ \cos(\frac{\pi}{4}) $. The cosine of $ \frac{\pi}{4} $ is $ \frac{\sqrt{2}}{2} $. Recall that $ \sec(x) = \frac{1}{\cos(x)} $, so:

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