化简:4.化简:(2cos²α-1)/{[2tan((π/4)- α)]}*{cos³[(π/4)- α]}
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![化简:4.化简:(2cos²α-1)/{[2tan((π/4)- α)]}*{cos³[(π/4)- α]}](/uploads/image/z/675160-16-0.jpg?t=%E5%8C%96%E7%AE%80%EF%BC%9A4.%E5%8C%96%E7%AE%80%EF%BC%9A%EF%BC%882cos%26%23178%3B%CE%B1-1%EF%BC%89%2F%7B%5B2tan%28%28%CF%80%2F4%29-+%CE%B1%29%5D%7D%2A%7Bcos%26%23179%3B%5B%28%CF%80%2F4%29-+%CE%B1%5D%7D)
化简:4.化简:(2cos²α-1)/{[2tan((π/4)- α)]}*{cos³[(π/4)- α]}
化简:4.化简:(2cos²α-1)/{[2tan((π/4)- α)]}*{cos³[(π/4)- α]}
化简:4.化简:(2cos²α-1)/{[2tan((π/4)- α)]}*{cos³[(π/4)- α]}
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)