若cos(π/4+x)cos(π/4-x)=1/4,则sinx的四次方+cosx的四次方=?
问题描述：
若cos(π/4+x)cos(π/4-x)=1/4,则sinx的四次方+cosx的四次方=?
答
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
答
cos^2x+cos^4x =1-sin^2x+(1-sin^2x)^2 =1-sin^2x+1-2sin^2x+=(1-sinx-1)(1-sinx-2) =-sinx(-sinx-1) =sinx(sinx+1) =sinx+
答
cos(a+b)=cosacosb-sinasinbcos(a-b)=cosacosb+sinasinbcos(a+b)+cos(a-b)=2cosacosbcos(π/4+x+π/4-x)+cos(π/4+x-π/4+x)=2cos(π/4+x)cos(π/4-x)=1/2cos(π/2)+cos(2x)=1/2cos2x=1/2sin2x=±√3/2sin²2x=3...