已知函数f(x)=(1-tanx)*[1+根号2sin(2x+π⼀4)]

2024-12-21 01:58:41
推荐回答(1个)
回答1:

(1)应是这样的分母≠0
当sin(π/2-x)= cosx = 0 X =(1/2 + K)PI,K∈Z
定义域:{ X | X(1/2 + K)裨∈R和x≠K∈Z}
(2)
简化原来的公式:
(x)的= [1 +方根2cos(2X-π/ 4)] /的罪(π/2-x)
=(1 + cos2x + sin2x)/ cosx
= 2(要写为sinx + cosx)
= 2√2sin(x +π/ 4)

当x∈[-π/ 4,π/ 2)
X +π/ 4∈[0,3π/ 4) BR /> SIN(X +π/ 4)∈[0,1]
∴F(X)∈[0,2√2]