1. ## inequalities of trigo

inequalities of trigo
p124 q12 re.ex5

2. Hello, afeasfaerw23231233!

We have to "talk" our way through these . .

Find the values of $\displaystyle \theta$ in the range $\displaystyle 0 \leq \theta < 2\pi$

$\displaystyle (a)\;\;2\sin^2\!\theta \:\leq \:1$

We have: .$\displaystyle \sin^2\!\theta \;\leq \;\frac{1}{2}\quad\Rightarrow\quad |\sin\theta| \;\leq \;\frac{1}{\sqrt{2}} \quad\Rightarrow\quad -\frac{1}{\sqrt{2}}\;\leq \;\sin\theta \;\leq \;\frac{1}{\sqrt{2}}$

Clearly, we are dealing with multiples of 45°: . $\displaystyle \frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{4},\: \frac{7\pi}{4}$

When is $\displaystyle \sin\theta$ between $\displaystyle -\frac{1}{\sqrt{2}} \text{ and } +\!\frac{1}{\sqrt{2}}$ ?

Examine the unit circle . . .

Answer: .$\displaystyle {\color{blue}\left[0,\:\frac{\pi}{4}\right],\;\left[\frac{3\pi}{4},\:\frac{5\pi}{4}\right], \;\left[\frac{7\pi}{4},\:2\pi\right)}$

$\displaystyle (b)\;\sin2\theta + \cos\theta \:\leq \:0$
This one takes a bit more "talking" . . .

We have: .$\displaystyle 2\sin\theta\cos\theta + \cos\theta \:\leq \:0 \quad\Rightarrow\quad \cos\theta(2\sin\theta + 1) \:\leq\:0$

The product of two factors is negative if the factors have opposite parity.
Hence, there are two cases to examine . . .

$\displaystyle [1]\;\begin{Bmatrix} \cos\theta & \leq & 0 \\ 2\sin\theta + 1 & \geq & 0 \end{Bmatrix}$

. . $\displaystyle \cos\theta$ is negative in quadrants 2 and 3: .$\displaystyle \left[\frac{\pi}{2},\:\frac{3\pi}{2}\right]$

. . $\displaystyle \sin\theta \:\geq \:-\frac{1}{2} \quad\Rightarrow\quad \left[0,\:\frac{7\pi}{6}\right] \cup \left[\frac{11\pi}{6},\:2\pi\right)$

The intersection of the two sets is: .$\displaystyle {\color{blue}\left[\frac{\pi}{2},\:\frac{7\pi}{6}\right]}$

$\displaystyle [2]\;\begin{Bmatrix}\cos\theta & \geq & 0 \\ 2\sin\theta + 1 & \leq 0 \end{Bmatrix}$

. . $\displaystyle \cos\theta$ is positive in quadrants 1 and 4: .$\displaystyle \left[0,\:\frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2},\:2\pi\right)$

. . $\displaystyle \sin\theta \:\leq\:-\frac{1}{2} \quad\Rightarrow\quad \left[\frac{7\pi}{6},\:\frac{11\pi}{6}\right]$

The intersection of the two sets is: .$\displaystyle {\color{blue}\left[\frac{7\pi}{6},\:\frac{3\pi}{2}\right]}$

The solution is: . $\displaystyle \left[\frac{\pi}{2},\:\frac{7\pi}{6}\right] \cup \left[\frac{7\pi}{6}\:\frac{3\pi}{2}\right] \;=\;{\bf{\color{blue}\left[\frac{\pi}{2},\:\frac{3\pi}{2}\right]}}$ . . . Quadrants 2 and 3