已知定义在R上的函数f(x)满足①:f(x+y)+f(x-y)=2f(x)cosy;②:f(0)=0,f(π/2)=1,求(1)判断f(x)的奇偶性;(2)求f(x)(3)求f(x)+cosx+f(x)cosx(4)求f(x)+cosx+f(x)·cosx的最大值.
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![已知定义在R上的函数f(x)满足①:f(x+y)+f(x-y)=2f(x)cosy;②:f(0)=0,f(π/2)=1,求(1)判断f(x)的奇偶性;(2)求f(x)(3)求f(x)+cosx+f(x)cosx(4)求f(x)+cosx+f(x)·cosx的最大值.](/uploads/image/z/6742880-8-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3%E2%91%A0%3Af%28x%2By%29%2Bf%28x-y%29%3D2f%28x%29cosy%3B%E2%91%A1%3Af%280%29%3D0%2Cf%28%CF%80%2F2%29%3D1%2C%E6%B1%82%281%29%E5%88%A4%E6%96%ADf%28x%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7%3B%282%29%E6%B1%82f%28x%29%283%29%E6%B1%82f%28x%29%2Bcosx%2Bf%28x%29cosx%284%29%E6%B1%82f%28x%29%2Bcosx%2Bf%28x%29%C2%B7cosx%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC.)
已知定义在R上的函数f(x)满足①:f(x+y)+f(x-y)=2f(x)cosy;②:f(0)=0,f(π/2)=1,求(1)判断f(x)的奇偶性;(2)求f(x)(3)求f(x)+cosx+f(x)cosx(4)求f(x)+cosx+f(x)·cosx的最大值.
已知定义在R上的函数f(x)满足①:f(x+y)+f(x-y)=2f(x)cosy;②:f(0)=0,f(π/2)=1,求
(1)判断f(x)的奇偶性;
(2)求f(x)
(3)求f(x)+cosx+f(x)cosx
(4)求f(x)+cosx+f(x)·cosx的最大值.
已知定义在R上的函数f(x)满足①:f(x+y)+f(x-y)=2f(x)cosy;②:f(0)=0,f(π/2)=1,求(1)判断f(x)的奇偶性;(2)求f(x)(3)求f(x)+cosx+f(x)cosx(4)求f(x)+cosx+f(x)·cosx的最大值.
(1) ①中令x=0,得 f(y)+f(-y)=0.故f(x)为奇函数.(2) 由①得 f(y+x)+f(y-x)=2f(y)cosx,故 f(x+y)-f(x-y)=2f(y)cosx,结合①得 f(x+y)= f(x)cosy+f(y)cosx,故 f(x+π/2) =f(π/2)cosx,f(x+π/2) =cosx,x-π/2代替x得f(x) =sinx,(3) f(x)+cosx+f(x)cosx= sinx+cosx+sinxcosx= sinx+cosx+(sin2x)/2.(4) 令g(x)=sinx+cosx+sinxcosx=√2∙sin(x+π/4)+(sin2x)/2,g(x-π/4)= √2∙sin(x)-(cos2x)/2,令t=sinx,则 g=t2+√2t-1/2,|t|≤1.t=1时 g取得最大值 √2+1/2,此时 x=π/2.
由①得f(x)+f(y)=2f[(x+y)/2]cos[(x-y)/2]
1、令y=-x得f(x)为奇函数。
2、对x求偏导,并令y=-x,得f'(x)=f'(0)cosx,于是f(x)=f'(0)sinx+C。而f(0)=0,f(π/2)=1得C=0,f'(0)=1.故f(x)=sinx
3、√2+1/2,x=π/4