Question

已知cos(a+π/4)=3/5, π/2≤a≤3π/2, 求cos(2a+π/4)的值

Answer 1

解：由1/2＜3/5＜√2/2且π/2≤α≤3π/2得
5π/3＜α+π/4＜7π/4，即17π/12＜α＜3π/2
cos(α+π/4)=cosαcos(π/4)-sinαsin(π/4)
=(√2/2)(cosα-sinα)=3/5
故cosα-sinα=3√2/5，平方得
cos²α-2cosαsinα+sin²α=18/25，即
sin2α=7/25，再由17π/6＜2α＜3π，得cos2α=-24/25，故
cos(2α+π/4)=cos2αcos(π/4)-sin2αsin(π/4)
=(√2/2)(cos2α-sin2α)=-31√2/50

Answer 2

π/2≤a≤3π/2
cos(a+π/4)=3/5>0
a+π/4>3π/2
3π/2<a+π/4≤7π/4
sin(a+π/4)=-√[1-cos²(a+π/4)]=-4/5
sina=[sin(a+π/4)-π/4]=√2/2*[sin(a+π/4)-cos(a+π/4)]=√2/2*(-4/5-3/5)=-7√2/10
cosa=-√(1-sin²a)=-√2/10
cos2a=2cos²a-1=-24/25,sin2a=2sinacosa=7/25
cos(2a+π/4)=√2/2(cos2a-sin2a)=-31√2/50

Answer 3

31√2/50