Suppose that $x$ is measured in radians. Find the maximum value of \[\frac{\sin2x+\sin4x+\sin6x}{\cos2x+\cos4x+\cos6x}\] for $0\le x\le \frac{\pi}{16}$
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Tags: trigonometry, algebra, function, domain
Dr Sonnhard Graubner
03.06.2014 19:10
hello, we obtain $\frac{\sin2x+\sin4x+\sin6x}{\cos2x+\cos4x+\cos6x} \le 1$ and the equal sign holds for $x=\frac{\pi}{16}$ Sonnhard.
Dukejukem
04.06.2014 04:07
Are you sure this is the exact problem? Each individual term of the numerator is clearly maximized when $x = \dfrac{\pi}{16}$, and similarly each individual term of the denominator is clearly minimized when $x = \dfrac{\pi}{16}$. Therefore the maximum occurs at $x = \dfrac{\pi}{16}$. We then have
$\dfrac{\sin 2x + \sin 4x + \sin 6x}{\cos 2x + \cos 4x + \cos 6x} = \dfrac{\sin \dfrac{\pi}{8} + \sin \dfrac{2\pi}{8} + \sin \dfrac{3\pi}{8}}{\cos \dfrac{\pi}{8} + \cos \dfrac{2\pi}{8} + \cos \dfrac{3\pi}{8}}$
$= \dfrac{\sin \dfrac{\pi}{8} + \sin \dfrac{2\pi}{8} + \sin \dfrac{3\pi}{8}}{\sin \dfrac{3\pi}{8} + \sin \dfrac{2\pi}{8} + \sin \dfrac{\pi}{8}} = 1$.
mcrasher
16.06.2014 17:07
SMOJ wrote: Suppose that $x$ is measured in radians. Find the maximum value of \[\frac{\sin2x+\sin4x+\sin6x}{\cos2x+\cos4x+\cos6x}\] for $0\le x\le \frac{\pi}{16}$ $\frac{\sin2x+\sin4x+\sin6x}{\cos2x+\cos4x+\cos6x}=\frac{2\sin 4x. \cos2x+\sin 4x}{2\cos 4x. \cos2x+\cos 4x}=\tan 2x\leq{1}$ ,for $0\le x\le \frac{\pi}{16}$