Geometry and Trigonometry - Trigonometric functions

Given that $\sin \theta = \frac{3}{5}$ and $\frac{\pi}{2} < \theta < \pi$, find the exact value of $\cos 2\theta$.

We use the double angle formula for cosine: $\cos 2\theta = 1 - 2\sin^{2}\theta$ Substitute the given value: $\cos 2\theta = 1 - 2\left(\frac{3}{5}\right)^{2}$ $\cos 2\theta = 1 - 2\left(\frac{9}{25}\right)$ $\cos 2\theta = 1 - \frac{18}{25}$ $\cos 2\theta = \frac{7}{25}$

Since we are given $\sin \theta$, the most direct formula to use is $\cos 2\theta = 1 - 2\sin^{2}\theta$. The interval $\frac{\pi}{2} < \theta < \pi$ indicates $\theta$ is in the second quadrant, but for the double angle cosine formula involving only $\sin^{2}\theta$, the quadrant of $\theta$ does not change the result because the sine value is squared.

Solve the equation $2\cos^{2}x + \sin x - 1 = 0$ for $0 \le x \le 2\pi$.

First, use the identity $\cos^{2}x = 1 - \sin^{2}x$ to get the equation in terms of sine: $2(1 - \sin^{2}x) + \sin x - 1 = 0$ $2 - 2\sin^{2}x + \sin x - 1 = 0$ $-2\sin^{2}x + \sin x + 1 = 0$ $2\sin^{2}x - \sin x - 1 = 0$ Factorizing the quadratic: $(2\sin x + 1)(\sin x - 1) = 0$ This gives $\sin x = -\frac{1}{2}$ or $\sin x = 1$. For $\sin x = 1$ in the domain: $x = \frac{\pi}{2}$. For $\sin x = -\frac{1}{2}$ in the domain: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ and $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$. Final solutions: $x \in \left\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\}$.

To solve an equation with both $\cos^{2}x$ and $\sin x$, we convert everything to $\sin x$ using the Pythagorean identity. This creates a quadratic equation in terms of $\sin x$, which can be solved by factoring or the quadratic formula.

A sector of a circle has a radius of $8$ cm and an arc length of $12$ cm. Find the area of the sector.

First, find the angle $\theta$ in radians using $l = r\theta$: $12 = 8\theta \implies \theta = \frac{12}{8} = 1.5 \text{ radians}$ Now, use the area formula $A = \frac{1}{2}r^{2}\theta$: $A = \frac{1}{2}(8)^{2}(1.5)$ $A = \frac{1}{2}(64)(1.5) = 32 \times 1.5 = 48$ The area is $48 \text{ cm}^{2}$.

The problem provides the radius and arc length. By using the arc length formula, we determine the central angle in radians, which is then plugged into the sector area formula.

1. The period of the function $h(t)$ is given by $T = \frac{2\pi}{b}$.
2. Here, $b = \frac{\pi}{10}$, so $T = \frac{2\pi}{\frac{\pi}{10}} = 2\pi \times \frac{10}{\pi} = 20$ minutes.
3. The maximum height occurs when $\cos\left(\frac{\pi}{10}t\right) = -1$.
4. $h_{\text{max}} = 15 - 12(-1) = 15 + 12 = 27$ meters.

The period $T$ represents the time for one cycle. In cosine functions of the form $d - a \cos(bt)$, the maximum value is $d + a$ and the minimum is $d - a$.

1. The reference angle (in the first quadrant) where $\tan x = \sqrt{3}$ is $x = \frac{\pi}{3}$.
2. Since $\tan x$ is positive in Quadrants I and III, we find the second solution.
3. In Quadrant III, $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$.
4. Thus, the solutions in the given range are $x = \frac{\pi}{3}$ and $x = \frac{4\pi}{3}$.

Tangent is positive in the first and third quadrants. We use the period of $\pi$ for tangent to find additional solutions within the domain.