Trig inequality

**livmed** · March 1st 2010, 01:53 PM

Although we haven't done them in school yet, I would like someone to show me how to solve it!!

$\cos^2 \frac{x}{3} \le \sin^2 \frac{x}{3} - \frac{1}{2}$

**Soroban** · March 2nd 2010, 06:30 PM

Hello, livmed!

$\cos^2\!\left(\frac{x}{3}\right) \;\le \;\sin^2\!\left(\frac{x}{3}\right) - \frac{1}{2}$

We have: . $\cos^2\!\left(\frac{x}{3}\right)- \sin^2\left(\frac{x}{3}\right) \;\leq \;-\frac{1}{2}$

. . Hence: . . . . . . . $\cos\left(\frac{2x}{3}\right) \;\leq\;-\frac{1}{2}$

We find that: . $\cos\theta\:\leq\:-\frac{1}{2}$ when $\theta$ is between $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$

So we have: . $\frac{2\pi}{3} \;\leq \;\frac{2x}{3} \;\leq\;\frac{4\pi}{3}$

Multiply by $\tfrac{3}{2}\!:\;\;\boxed{\pi \;\leq \;x \;\leq \;2\pi }$

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