Question

已知cos(π/4+a)cos(π/4-a)=1/4 求sina的四次方加cosa的四次方 求助啊。。。。。。

Answer 1

cos（π/4-a）=sin[π/2-(π/4-a)]=sin(π/4+a)
cos(π/4+a)cos(π/4-a)

=cos(π/4+a)sin(π/4+a)
=1/2 sin(π/2+ 2a)
=1/2cos2a= 1/4
即 cos2a= 1/2
sin²2a+cos²2a=1
即sin²2a= 3/4
sina的四次方加cosa的四次方

=(sin²a+cos²a)²- 2sin²acosa²
=1-[1/2sin²2a]
=5/8
如有不明白，可以追问如有帮助，记得采纳，谢谢

Answer 2

解答：
cos(π/4+a)=cos[π/2-(π/4-a)]=sin(π/4-a)
∵cos(π/4+a)cos(π/4-a)=1/4
∴ sin(π/4-a)cos(π/4-a)=1/4
(1/2)sin(π/2-2a)=1/4
∴ (1/2)cos(2a)=1/4
cos(2a)=1/2
sin²(2a)=1-cos²(2a)=1-1/4=3/4
sina的四次方加cosa的四次方
=(sin²a+cos²a)-2sin²acos²a
=1-(1/2)(2sinacosa)²
=1-(1/2)(sin2a)²
=1-(1/2)*3/4
=1-3/8
=5/8

Answer 3

cos(π/4+a)cos(π/4-a)=1/4
cos(π/4+a)cos(π/4-a)=1/2[cos(π/4+a+π/4-a)+cos(π/4+a-π/4+a)]
=1/2[cos(π/2)+cos2a]=1/4
cos2a=1/2
sina^4+cosa^4=(sina^2+cosa^2)^4-2sina^2+cosa^2
=1-1/2(sin2a)^2=1-1/2[1-(cos2a)^2]=1-1/2[1-1/4]=5/8