Question

高一数学必修4三角函数

cos²(π/4-α)+cos²(π/4+α)=

cos²((4n+1/4)π+α)+cos²((4n-1/4)π-α) （n∈Z）

Answer 1

1.原式＝（cosπ/4cosα sinπ/4sinα)^2 (cosπ/4cosα-sinπ/4sinα)^2=1/2(cos^2α sin^2α 2cosαsinα) 1/2(cos^2α sin^2α－2cosαsinα)=1/2 1/2=1第二题同理化简展开也等于1

Answer 2

sin(π/2+α）=cosα
1 cos²(π/4-α)=sin²(π/4-α) cos²(π/4-α)+cos²(π/4+α)=1
2 cos²((4n+1/4)π+α)+cos²((4n-1/4)π-α)
=cos²(1/4π+α)+cos²(1/4)π-α)
=sin²α+cos²α=1

Answer 3

考 这么简单还拿出来晒