图
5.sin(π/4-a)=5/13,
所以(cosa)^2-(sina)^2
=cos2a
=sin(π/2-2a)
=2sin(π/4-a)cos(π/4-a),
所以原式=2sin(π/4-a)=10/13.
sin(π/4-x) =5/13
To find [ (cosx)^2 - (sinx)^2]/ cos(π/4-x)
solution:
sin(π/4-x) =5/13
(√2/2)( cosx - sinx ) =5/13
(1/2)( 1- sin2x ) = 5/13
sin2x = 3/13
[ (cosx)^2 - (sinx)^2]/ cos(π/4-x)
= cos2x/cos(π/4-x)
=[√(13^2-3^2)/13 ]/[√(13^2-5^2)/13]
= 4√10/12
=√10/3
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