f(x)=3sinx+2cosx
(bcosc)/a = -1,详见下面:
af(x)+bf(x-c)=a(3sinx+2cosx+1)+b[3sin(x-c)+2cos(x-c)+1]=1
上式对一切x成立,
取x=0,得到:3a-3bsinc+2bcosc+b=1……(1)
取x=π/2,得到:4a+3bcosc+2bsinc+b=1……(2)
取x=π,得到:-a+3bsinc-2bcosc+b=1……(3)
2*(1)+3*(2),得到:18a+13bcosc+5b=5……(4)
(1)+(3),得到:2a+2b=2 ==> a+b=1,
(4)写成:13a+13bcosc+5(a+b)=5,以a+b=1代入,得到...全部
(bcosc)/a = -1,详见下面:
af(x)+bf(x-c)=a(3sinx+2cosx+1)+b[3sin(x-c)+2cos(x-c)+1]=1
上式对一切x成立,
取x=0,得到:3a-3bsinc+2bcosc+b=1……(1)
取x=π/2,得到:4a+3bcosc+2bsinc+b=1……(2)
取x=π,得到:-a+3bsinc-2bcosc+b=1……(3)
2*(1)+3*(2),得到:18a+13bcosc+5b=5……(4)
(1)+(3),得到:2a+2b=2 ==> a+b=1,
(4)写成:13a+13bcosc+5(a+b)=5,以a+b=1代入,得到:
13a+13bcosc=0 ==> a+bcosc=0 ==> bcosc=-a ==> (bcosc)/a=-1。
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