来自罗绍基的问题
【（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】
（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!
1回答
2020-05-09 00:10