计算log2sinπ/12+log2cosπ/12的值为
问题描述:
计算log2sinπ/12+log2cosπ/12的值为
log2sinπ/12+log2cosπ/12
=log2(sinπ/12cosπ/12)
=log2(1/2sinπ/6)
=log2(1/4)
=-2第一步是怎么来的?
答
公式:loga(bc)=loga(b)+loga(c)那第二步呢公式:sin2A=2sinA/2*cosA/2sinπ/12cosπ/12=1/2*(2*sinπ/12cosπ/12)=1/2sinπ/6