Question

若cos(π/4+x)cos(π/4-x)=1/4,则sinx的四次方+cosx的四次方=？

Answer 1

cos(π/4+x)cos(π/4-x)=1/2(CoS(2X)+COS(1/2π))=1/4,
CoS2X=1/2
SINX^4+CoxX^4=(sINX^2+CosX^2)^2-2(SINX*COsX)^2=1-1/2*Sin2X^2=1-1/2(1-Cos^2X)=1-1/2*3/4=5/8

Answer 2

cos(a+b)=cosacosb-sinasinb
cos(a-b)=cosacosb+sinasinb
cos(a+b)+cos(a-b)=2cosacosb
cos(π/4+x+π/4-x)+cos(π/4+x-π/4+x)=2cos(π/4+x)cos(π/4-x)=1/2
cos(π/2)+cos(2x)=1/2
cos2x=1/2
sin2x=±√3/2
sin²2x=3/4
(sin²x+cos²x)²
=1
=sinx^4+cosx^4+2sin²xcos²x
=sinx^4+cosx^4+1/2 * 3/4
=sinx^4+cosx^4+3/8
sinx^4+cosx^4=5/8