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[不等式] 2元的三角范围问题

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力工 posted 2020-5-11 08:56 |Read mode
大神们,这题反正我是蒙圈了。就象遇到了“吸星大法”,不知道打出去后的方向,
到底怎么求呢?求各路大神您帮助。
已知$x,y$满足$2cosx+3cosy=6cos(x-y)+1$,求$cos(x-y)$的取值范围。

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kuing posted 2020-5-11 18:06
不难啊……令 `x=y+t`,则等式化为
\begin{align*}
6\cos t+1&=2\cos(y+t)+3\cos y\\
&=(2\cos t+3)\cos y-2\sin t\sin y\\
&=\sqrt{(2\cos t+3)^2+4\sin^2t}\sin(y+\varphi)\\
&=\sqrt{13+12\cos t}\sin(y+\varphi),
\end{align*}所以
\[(6\cos t+1)^2\leqslant13+12\cos t,\]解得 `-\sqrt{1/3}\leqslant\cos t\leqslant\sqrt{1/3}`。

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original poster 力工 posted 2020-5-11 19:50
回复 2# kuing
一个字!
服!渣猹没有方向,所以豺狼啊。

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