数学涵数

f(x)=1/2cos2x+1/2+1/2sin2x (cos2x=2cos^2(x)-1)
f（x）=（根号2/2)sin(2X+π/4)+(1/2)

(1)f(3π/8)=1/2

2.f(x)=√2/2cos(2x-π/4)+1/2
cosx函数在[(2k-1)π,2kπ](k∈Z)是增函数
所以 (2k-1)π≤2x-π/4≤2kπ
(k-3/8)π≤x≤（k+1/8)π

f(x)=(cos X)^2+sin X*cos
=（1/2)(cos2x+1）+(1/2)sin2x
=(1/2)(cos2x+1+sin2x)
f(3π/8)=(1/2)(cos2(3π/8)+1+sin2(3π/8))
=(1/2)(cos(3π/4)+1+sin(3π/4))
=(1/2)(cos(3π/4)+1+sin(3π/4))
因为 cos(3π/4) = -sin(3π/4)
所以cos(3π/4) + sin(3π/4)=0
所以f(3π/8)=1/2

2.f(x)=√2/2cos(2x-π/4)+1/2
cosx函数在[(2k-1)π,2kπ](k∈Z)是增函数
所以 (2k-1)π≤2x-π/4≤2kπ
(k-3/8)π≤x≤（k+1/8)π

另外
函数的"函"字你打错了

f(x)=(cos X)^2+sin X*cos
=（1/2)(cos2x+1）+(1/2)sin2x
=(1/2)(cos2x+1+sin2x)
f(3π/8)=(1/2)(cos2(3π/8)+1+sin2(3π/8))
=(1/2)(cos(3π/4)+1+sin(3π/4))
=(1/2)(cos(3π/4)+1+sin(3π/4))
因为 cos(3π/4) = -sin(3π/4)
所以cos(3π/4) + sin(3π/4)=0