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kuing
Posted 2017-1-22 15:50
由积化和差有
\begin{align*}
&\sin x\sin \left( \frac{5\pi }{12}+x \right)
+\sin \left( \frac\pi3-x \right)\sin \left( \frac{3\pi }4-x \right)\\
={}&\frac12\left( \cos \frac{5\pi }{12}-\cos \left( \frac{5\pi }{12}+2x \right) \right)
+\frac12\left( \cos \frac{5\pi }{12}-\cos \left( \frac{13\pi }{12}-2x \right) \right) \\
={}&\sin \frac\pi{12}-\frac12\sin \left( \frac\pi{12}-2x \right)
+\frac12\cos \left( \frac\pi{12}-2x \right)\\
={}&\frac{\sqrt6-\sqrt2}4-\frac{\sqrt2}2\cos \left( \frac\pi3-2x \right),
\end{align*}
且
\[
\sin \left( \frac{3\pi }4-x \right)\sin \left( \frac{5\pi }{12}+x \right)
=\frac12\left( \cos \left( \frac\pi3-2x \right)+\frac{\sqrt3}{2} \right),
\]
故令 $t=\cos(\pi/3-2x)\in[1/2,1]$，则
\[f(x)=\frac{\sqrt6-\sqrt2+2\sqrt2t}{2t+\sqrt3}
=\sqrt2-\frac{\sqrt2}{2t+\sqrt3}
\in\left[\frac{\sqrt6}{1+\sqrt3},\frac{\sqrt6+\sqrt2}{2+\sqrt3}\right]
=\left[\frac{3\sqrt2-\sqrt6}2,\sqrt6-\sqrt2\right].\] |
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游客
Posted 2017-2-22 14:34
