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[函数] 求三角式的值

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lemondian Posted 2020-1-30 20:20 |Read mode
求三角式的值:
$cos\dfrac{2\pi}{7}+2cos\dfrac{4\pi}{7}+3cos\dfrac{6\pi}{7}+4cos\dfrac{8\pi}{7}+5cos\dfrac{10\pi}{7}+6cos\dfrac{12\pi}{7}$.
另外:
(1)能否推广?
(2)正弦能求吗?

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kuing Posted 2020-1-30 22:24
原题超简单吧,化简就是 `7(\cos(2\pi/7)+\cos(4\pi/7)+\cos(6\pi/7))`,别说这个也不会。

推广的话看你想往哪方向推了,是想改分母,还是改系数或其他东西,自己玩。

改正弦的话就原题都不行。

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