Question

已知cos（x+π/4)=3/5且17π/12＜x＜7π/4，求（sin2x+2sin^2x）/ （1-tanx）的值域

Answer 1

cos（x+π/4)=3/5
=> cos x cos π/4- sinx sinπ/4 = 3/5
=> cos x -sin x = 3(2)^(1/2) / 5 ...... (1)
=> (cos x)^2 - 2 cosx sinx + (sinx)^2 = 18/25
=> 2cosx sinx = 7/25 ........(2)
=> (cos x)^ 2+2cosx sinx + (sinx)^2 = 18/25+4 2sinxcosx = 18/25+28/25= 46/25
=> cosx + sinx = +/-((46)^(1/2)/5)
而 （sin2x+2sin^2x）/ （1-tanx）= [2sin x cos x + 2 (sin x)^2] / [1-tan x]
= 2 sin x cos x [sin x + cos x ]/ [cos x -sin x]
= 2 (7/25) (+/-((46)^(1/2)/5))(3(2)^(1/2) / 25)
= +/-[ 42(46)^(1/2)/ 3125]