(sinx+cosx)^2+2sin^2(π/4-x)
1个回答
(sinx+cosx)^2+2sin^2(π/4-x)
=(1+sin2x)+1-cos2(π/4-x)
=2+sin2x-sin2x=2
相关问题
2sin(x-π/4)sin(x+π/4)=(sinx-cosx)(sinx+cosx)=-cos2a
f(x)=4sin^2[(π+2x)/4].sinx+(cosx+sinx)(cosx-sinx)
化简f(x)=4sinx*sin^2((π+2x)/4)+(cosx+sinx)(cosx-sinx)
求解:f(x)=2sinx*(π/2-x)-√3sin(π+x)cosx+sin(π/2+x)cosx
【(1+sinx)/cosx】*(sin2x/(2cos^2(π/4-x/2)))
y=sinx+cosx可化成_____?A.2sin(x-π/4);B.2sin(x+π/4); C.√2sin(x-π
若(cos2x)/[sin(x-π/4)]=(-根号2)/2,则cosx+sinx=?
化简:(1)3/2cosx-√3/2sinx(2)√3sinx/2+cosx/2(3)√2/4sin(π/4-x)+√6
sinx+cosx+sinxcosx+1=√2*sin(x+π/4)+1/2 *sin2x+1是为什么?
化简:(1)根三sinx+cosx (2)sinx-cosx (3)3sinx+4cosx (4)根二sin(π/4-x