cos(π/6-α)=m,m绝对值小于等于1,求cos(5π/6+α)
问题描述：
cos(π/6-α)=m,m绝对值小于等于1,求cos(5π/6+α)
答
cos(π/6-α)=cos[π-(5π/6+α)]=-cos(5π/6+α)
所以cos(5π/6+α)=-m
记住奇变偶不变 符号看象限
答
cosπ=-1
=cos[(π/6-a)+(5π/6+a)]
=cos(π/6-a)cos(5π/6+a)-sin(π/6-a)sin(5π/6+a)
=mcos(5π/6+a)-根号(1-m^2)根号mcos(5π/6+a）
设mcos(5π/6+a）=x
mx-根号（1-m^2）根号1-x^2
=-1
解出x=-m
答
cos(5π/6+α)=cos[π-(π/6-α)]=-cos[-(π/6-α)]
=-cos(π/6-α)=-m
答
cos(5π/6+α)=cos[π-(π/6-α)]
=cosπcos(π/6-α)+sinπsin(π/6-α)
=-1*m+0*根号下1-m2
=-m
答
cos(5π/6+α)=-cos[π-（5π/6+α)]=-cos(π/6-α)
所以cos(5π/6+α)=-m