求证cosx+sinx=√2cos(x-π/4)
1个回答
左边=√2(√2/2*cosx+√2/2*sinx)
=√2(cosxcosπ/4+sinπ/4sinπ/4)
=√2cos(x-π/4)
=右边
命题得证
相关问题
2sin(x-π/4)sin(x+π/4)=(sinx-cosx)(sinx+cosx)=-cos2a
求证:tan(x-π/4)=(sinx-cosx)/(sinx+cosx)
求证:cosx+根3sinx=2cos(x-π/3)
cosx-sinx=根号2*cos(x+4/π)吗
y=(cos2(x-π/4)-1/2)/(1+sinx+cosx) (0≤x≤π/2),设t=sinx+cosx 1)求
已知函数y=(cos(x-(π/4))-0.5)/(1+sinx+cosx),0≤x≤π/2,设t=sinx+cosx
f(x)=4sin^2[(π+2x)/4].sinx+(cosx+sinx)(cosx-sinx)
【(1+sinx)/cosx】*(sin2x/(2cos^2(π/4-x/2)))
已知函数f(x)=cos(2x-π/3)+(cosx+sinx)(cosx-sinx)
4cosx ( 1/2 cosx - √3/2 sinx ) = 4cosx cos( x + π/3 )是怎么来的