Question

如果cos(π/4+a)cos(π/4-a)=1/4 求sina的四次方加cosa的四次方

Answer 1

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 2

cos(π/4+a)cos(π/4-a)=1/4
cos(π/4+a)cos(π/2-π/4-a)=1/4
cos(π/4+a)cos[π/2-(π/4+a)]=1/4
cos(π/4+a)sin(π/4+a)=1/4
sin(π/4+a)cos(π/4+a)=1/4
2sin(π/4+a)cos(π/4+a)=1/2
sin(π/2+2a)=1/2
cos2a=1/2
cos^2a-sin^2a=1/2

sin^4a+cos^4a
=sin^4a-2sin^2acos^2a+cos^4a+2sin^2acos^2a
=(cos^2a-sin^2a)^2+2sin^2acos^2a
=(1/2)^2+2sin^2acos^2a

=1/4+2sin^2acos^2a
=1/4+1/2*(sin2a)^2
=1/4+1/2*[1-(cos2a)^2]
=1/4+1/2*[1-1/4]
=1/4+1/2*3/4
=1/4+3/8
=2/8+3/8
=5/8

Answer 3

解：∵cos(π/4+a)cos(π/4-a)=1/4
∴(cosπ/4cosa-sinπ/4sina)(cosπ/4cosa+sinπ/4sina)=1/4
∴(cosπ/4cosa)^2-(sinπ/4sina)^2=1/4
∵(cosπ/4)^2=(sinπ/4)^2=1/2
∴(cosa)^2-(sina)^2=1/2
又∵(cosa)^2+(sina)^2=1
∴(cosa)^2=3/4，(sina)^2=1/4
∴(sina)^4+(cosa)^4=(1/4)^2+(3/4)^2=5/8
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