f（x）=cos^x/2— sin^x/2+sinx 当xo∈（0,π/4）且f（x0）=4根2/5,求f（x0+π/6）的值
问题描述：
f（x）=cos^x/2— sin^x/2+sinx 当xo∈（0,π/4）且f（x0）=4根2/5,求f（x0+π/6）的值
答
f(x)=(cos(x/2))^2 /2 -(sin(x/2))^2+sinx=cosx+sinx=√2sin(x+π/4 )f(x0)= √2sin(x0+π/4 )=4√2/5sin(x0+π/4)=4/5 cos(x0+π/4)=3/5f(x0+π/6)=sin[(x0+π/6)+π/4]=sin[(x0+π/4)+π/6]=(4/5)(√3/2)+(3/5)*(1...为什么f(x)=(cos(x/2))^2 /2 -(sin(x/2))^2+sinx =cosx+sinx我算的是= 1+cosx/2+1-cosx/2+sinx=1+cosx-1+cosx+2sinx=2cosx+2sinx不知怎么搞的~请问我那个想错了？？急急急急~~~~~~~~~~~~cos2a=(cosa)^2-(sina)^2=2(cosa)^2-1f(x)=(1+cosx)/2-(1-cosx)/2+sinx=1/2+cosx-1/2+sinx=cosx+sinx