sin^4pai/8+cos^4pai/8求值

sin^2a=[1-cos(2a)]/2
con^2a=[1+cos(2a)]/2
原式=[（1-cos pai/4）/2]^2+[(1+cos pai/4)/2]^2
=3/4

[sin(π/8)]^4+[cos(π/8)]^4
=[sin(π/8)]^4+[cos(π/8)]^2-2[sin(π/8)]^2*[cos(π/8)]^2
=1-(1/2)[2sin(π/8)*cos(π/8)]^2
=1-(1/2)[sin(π/4)]^2
=1-(1/2)[√2/2]^2
=3/4

[sin(π/8)]^4+[cos(π/8)]^4
={[sin(π/8)]^2+[cos(π/8)]^2}^2-2[sin(π/8)]^2[cos(π/8)]^2
=1^2-2[sin(π/8)cos(π/8)]^2
=1-2[sin(π/4)/2]^2
=1-sin(π/4)^2/2
=1-(√2/2)^2/2
=1-(1/2)/2
=1-(1/4)
=3/4.

sin^4pai/8+cos^4pai/8=（sin^2pai/8+cos^2pai/8）-2sin pai/8+cos pai/8=1-sin pai/4=1-sqrt2/2