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求值,三角函数
log2(cos20)+log2(cos40)+log2(cos80)
人气:251 ℃ 时间:2020-06-08 18:45:13
解答
原式=log2(cos20cos40cos80)
=log2(2sin20cos20cos40cos80/2sin20)
=log2(sin40cos40cos80/2sin20)
=log2(sin80cos80/4sin20)
=log2(sin160/8sin20)
=log2(sin20/8sin20)
=log2(1/8)
=-3
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