Assistance with trigonometric inequality needed please

**Squared** · February 16th 2011, 10:01 AM

Hey everyone, I'm doing some inequalities and I'm stuck with some trigonometric ones as I've never really encountered them before, some help with the following problem will be greatly appreciated.

Determine Θ, 0<=Θ<=2π so that (π = pi)

$2 + 3sin-cos2>=0$

What I tried was to replace cos2Θ with $1 - 2sin^2$ and factorize till I end up with
$(2sin + 1)(sin +1) >= 0$

From there on I'm not certain. I don't want to sinΘ >= -0.5 etc. because I'm afraid that will probably be wrong, How do I proceed? Thanks so much!

**skeeter** · February 16th 2011, 04:33 PM

Originally Posted by Squared

Hey everyone, I'm doing some inequalities and I'm stuck with some trigonometric ones as I've never really encountered them before, some help with the following problem will be greatly appreciated.

Determine Θ, 0<=Θ<=2π so that (π = pi)

$2 + 3sin-cos2>=0$

What I tried was to replace cos2Θ with $1 - 2sin^2$ and factorize till I end up with
$(2sin + 1)(sin +1) >= 0$

From there on I'm not certain. I don't want to sinΘ >= -0.5 etc. because I'm afraid that will probably be wrong, How do I proceed? Thanks so much!

$(2\sin{t} + 1)(\sin{t} +1) \ge 0$

the left side equals 0 at $t = \dfrac{7\pi}{6}$ , $\dfrac{11\pi}{6}$ , and $\dfrac{3\pi}{2}$

mark these three angles on a number line between $0$ and $2\pi$ . check the value of the left side of the inequality with an angle selected from each interval between the marks ... you're looking for the interval(s) that yields a positive value.

**Prove It** · February 16th 2011, 05:26 PM

I assume this should read

$\displaystyle 2 + 3\sin{\theta} - \cos{2\theta} \geq 0$ .

Use the identity $\displaystyle \cos{2\theta} \equiv 1 - 2\sin^2{\theta}$ and this inequality becomes

$\displaystyle 2 + 3\sin{\theta} - (1 - 2\sin^2{\theta}) \geq 0$

$\displaystyle 2\sin^2{\theta} + 3\sin{\theta} + 1 \geq 0$

Let $\displaystyle x = \sin{\theta}$ and the inequality can be written as

$\displaystyle 2x^2 + 3x + 1 \geq 0$ , a Quadratic inequation. Quadratic inequalities are almost always easiest to solve by completing the square.

$\displaystyle x^2 + \frac{3}{2}x + \frac{1}{2} \geq 0$

$\displaystyle x^2 + \frac{3}{2}x + \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 + \frac{1}{2} \geq 0$

$\displaystyle \left(x + \frac{3}{4}\right)^2 - \frac{9}{16} + \frac{8}{16} \geq 0$

$\displaystyle \left(x + \frac{3}{4}\right)^2 - \frac{1}{16} \geq 0$

$\displaystyle \left(x + \frac{3}{4}\right)^2 \geq \frac{1}{16}$

$\displaystyle \left|x + \frac{3}{4}\right| \geq \frac{1}{4}$

$\displaystyle x + \frac{3}{4} \leq -\frac{1}{4}$ or $\displaystyle x + \frac{3}{4} \geq \frac{1}{4}$

$\displaystyle x \leq -1$ or $\displaystyle x \geq -\frac{1}{2}$ .

$\displaystyle \sin{\theta} \leq -1$ or $\displaystyle \sin{\theta} \geq -\frac{1}{2}$ .

Case 1: Since $\displaystyle -1 \leq \sin{\theta} \leq 1$ for all real $\displaystyle \theta$ , that means the inequality $\displaystyle \sin{\theta} \leq -1$ is only true when $\displaystyle \sin{\theta} = -1$ .

$\displaystyle \theta = \frac{3\pi}{2}$ .

Case 2: $\displaystyle \sin{\theta} \geq -\frac{1}{2}$ .

First, we can solve where $\displaystyle \sin{\theta} = -\frac{1}{2}$

$\displaystyle \theta = \left\{\pi + \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\} = \left\{ \frac{7\pi}{6}, \frac{11\pi}{6}\right\}$ .

It should be clear that in the third quadrant of the unit circle, as your angle increases, the vertical distance gets greater in magnitude (so more negative). So any $\displaystyle \theta > \frac{7\pi}{6}$ will make $\displaystyle \sin{\theta} < -\frac{1}{2}$ . Also, in the fourth quadrant, as your angle increases, the vertical distance gets smaller in magnitude (so less negative). So any $\displaystyle \theta < \frac{11\pi}{6}$ will have $\displaystyle \sin{\theta} < -\frac{1}{2}$ . These are the values we disregard.

So finally, we have the solution $\displaystyle \theta \in \left[0, \frac{7\pi}{6}\right] \cup \left\{\frac{3\pi}{2}\right\} \cup \left[\frac{11\pi}{6}, 2\pi\right]$ .

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