求值log2sin(π/8)+log2cos(π/8)
问题描述：
求值log2sin(π/8)+log2cos(π/8)
2是以2为底
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