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求值log2sin(π/8)+log2cos(π/8)
2是以2为底
人气:218 ℃ 时间:2020-07-11 02:49:19
解答
log2sin(π/8)+log2cos(π/8)
=log2[sin(π/8)*cos(π/8)]
=log2[sin(π/4)/2]
=log2sin(π/4)-log2 2
=log2(√2/2)-1
=log2 2^(-1/2)-1
=-1/2-1
=-3/2
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