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[函数] `\tan 3\theta=\tan \theta \tan 2\theta\tan 4\theta`

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isee Posted 2021-7-8 23:20 |Read mode
若`\theta`为锐角满足`\tan 3\theta=\tan\theta\tan 2\theta\tan 4\theta`,则`\theta=`______________.

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 Author| isee Posted 2021-7-8 23:27
知乎上看到的,是一道选择题,这里把选项删除了。

很久不见三角函数中3倍角公式了,熟知`\tan 3\theta=\tan (\pi/3-\theta)\tan\theta\tan (\pi/3+\theta)`,结合已知有\[6\theta=\pm\frac {2\pi}3+k\pi,k\in \mathrm Z,\]于是\[\theta=\frac {\pi}9,\frac {4\pi}9,\frac {2\pi}9.\]

PS:注意:结果过程有大幅省略,不严谨

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kuing Posted 2021-7-9 02:53
切化弦
\[\sin3x\cos x\cos2x\cos4x=\cos3x\sin x\sin2x\sin4x,\]两两积化和差
\[(\sin4x+\sin2x)(\cos6x+\cos2x)=(\sin4x-\sin2x)(-\cos6x+\cos2x),\]展开化简为
\[\sin4x\cos6x+\sin2x\cos2x=0,\]即
\[\sin4x\left( \cos6x+\frac12 \right)=0,\]如果 `\sin4x=0`,由 `x` 为锐角,只能 `x=45\du`,但原式有 `\tan2x`,不允许,所以只剩下 `\cos6x=-1/2`,得到 `6x=120\du`, `240\du`, `480\du`,即 `x=20\du`, `40\du`, `80\du`。

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