sina+cosa=1 ,求sin^4(a)+cos^4(a)+sin(360°-a)cos(360°-a)的值如题
网友回答
sin^4(a)+cos^4(a)+sin(360°-a)cos(360°-a)
=sin^4(a)+cos^4(a)+sin(-a)cos(-a)
=sin^4(a)+cos^4(a)+sin(a)cos(a)
=[sin^2(a)+cos^2(a)]^2-2sin^2(a)cos^2(a)+sin(a)cos(a)
=1-2sin^2(a)cos^2(a)+sin(a)cos(a)
sina+cosa=1
两边平方得sin^2(a)+cos^2(a)+2sin(a)cos(a)=1
则1+2sin(a)cos(a)=1
则sin(a)cos(a)=0
所以1-2sin^2(a)cos^2(a)+sin(a)cos(a)=1
即sin^4(a)+cos^4(a)+sin(360°-a)cos(360°-a)=1