Question

已知函数f(x)={1-[(√2)sin(2x-π/4)]}/cosx 详细解题过程，谢谢！

（1）求f(x)的定义域
（2）设α是第四象限的角，且tanα=-4/3，求f(α)的值.

Answer 1

解：(1)cosx≠0
x≠Kπ+π/2 K∈Z
(2)由tanα=-4/3可知:sinα=4/5,cosα=-3/5
得：sin2α=2sinα*cosα=-24/25
cos2α=1-2(sinα)^2=-7/25
f(α)=[1-√2sin(2α-π/4)]/cosα
=[1-√2(sin2α*cosπ/4-cos2α*sinπ/4)]/cosα
=[1-sin2α+cos2α]/cosα
=[1+24/25-7/25]/(-3/5)
=-14/5

Answer 2

(1)
f(x)={1-[(√2)sin(2x-π/4)]}/cosx

f(x) define over real number except
cosx = 0
x = (2k+1)π/2 , k =0,±1,±2,....
f(x)的定义域
= R / A
where A = {(2k+1)π/2 , k =0,±1,±2,....}
(2)

tanα=-4/3
sinα = -4/5
cosα = 3/5
f(α))={1-[(√2)sin(2α-π/4)]}/cosα
= {1- (1/2)( sin2α - cos2α)} / cosα
= { 1- (1/2)( 2sinαcocosα - [cosα]^2+ [sinα]^2 } / cosα
= { 1- (1/2) ( 2(-12/25) - 9/25 + 16/25 }/ (3/5)
= {1+ (1/2) (17/25) } /(3/5)
= (67/50) ( 5/3)
= 201/10