（二倍角的三角函数）
问题描述：
（二倍角的三角函数）
cosπ/5cos2/5π的值是______
cosπ/5cos2π/5
=sinπ/5cosπ/5cos2π/5/sinπ/5
=4sinπ/5cosπ/5cos2π/5/4sinπ/5
=2sin2π/5cos2π/5/4sinπ/5
=sin4π/5/4sinπ/5
=sin(π-π/5)/4sinπ/5
=(1/4)sinπ/5/sinπ/5
=1/4
但从4sinπ/5cosπ/5cos2π/5/4sinπ/5
=2sin2π/5cos2π/5/4sinπ/5
=sin4π/5/4sinπ/5
=sin(π-π/5)/4sinπ/5
=(1/4)sinπ/5/sinπ/5
=1/4开始没看懂!不知道是怎么将式子变形的!
答
由公式：sin(a+b)=sina*cosb+sinb*cosa可得：sin(π/5 + π/5) = 2sin(π/5)*cos(π/5).4sin(π/5)*cos(π/5) = 2sin(2π/5);2sin(2π/5)*cos(2π/5)=sin(4π/5);4sin(π/5)*cos(π/5)*cos(2π/5)/(4sinπ/5)=2sin(2...