由于：
cos^3[π/4-a]
=[cos(π/4-a)]^3
=[cos(π/4)cosa+sin(π/4)sina]^3
=[(√2/2)cosa+(√2/2)sina]^3
=(√2/4)(cosa+sina)^3
tan[(π/4)- a]
=[tan(π/4)-tana]/[1+tan(π/4)tana]
=[1-tana]/[1+tana]
=[cosa-sina]/[cosa+sina]
所以原式
=[2cos^2(a)-1]/{[(2cosa-2sina)/(sina+cosa)]*(√2/4)(cosa+sina)^3}
=[2cos^2(a)-1]/[(√2/2)(cosa-sina)(cosa+sina)^2]
=(√2)[2cos^2(a)-1]/[(cosa-sina)(sina+cosa)(sina+cosa)]
=[√2cos2a]/[(cos^2(a)-sin^2(a))(sina+cosa)]
=[√2cos2a]/[cos2a(sina+cosa)]
=√2/(sina+cosa)