0＜a＜π/4.sin(π/4-a)＝5/13.求cos2a/cos(π/4+a)

0＜a＜π/4.sin(π/4-a)＝5/13.求cos2a/cos(π/4+a)
0＜a＜π/4.sin(π/4-a)＝5/13.求cos2a/cos(π/4+a)

0＜a＜π/4.sin(π/4-a)＝5/13.求cos2a/cos(π/4+a)

cos2a/cos(π/4+a)
=sin(π/2+2a)/cos(π/4+a)
=2sin(π/4+a)*cos(π/4+a)/cos(π/4+a)
=2sin(π/4+a)
=2cos[π/2-(π/4+a)]
=2cos(π/4-a)
因为 0＜a＜π/4, 0<π/4-a<π/4
所以 2cos(π/4-a)=2√[1-sin²(π/4-a)]=2√(1-25/169)=2*12/13=24/13