求证:1/(sinˆ2xcosˆ2x)—2=2(3+cos4x)/(1—cos4x)
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![求证:1/(sinˆ2xcosˆ2x)—2=2(3+cos4x)/(1—cos4x)](/uploads/image/z/11461383-63-3.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9A1%2F%28sin%26%23710%3B2xcos%26%23710%3B2x%29%E2%80%942%3D2%EF%BC%883%2Bcos4x%EF%BC%89%2F%281%E2%80%94cos4x%29)
求证:1/(sinˆ2xcosˆ2x)—2=2(3+cos4x)/(1—cos4x)
求证:1/(sinˆ2xcosˆ2x)—2=2(3+cos4x)/(1—cos4x)
求证:1/(sinˆ2xcosˆ2x)—2=2(3+cos4x)/(1—cos4x)
证明:
右边=2(3+cos4x)/(1—cos4x)
=(6+2cos4x)/(1—cos4x)
=(6+2cos4x)/[1-(1-2sin2x^2)]
=(6+2cos4x)/(1-1+2sin2x^2)
=(6+2cos4x)/2sin2x^2
=(3+cos4x)/sin2x^2
=(3+1-2sin2x^2)/sin2x^2
=(4-2sin2x^2)/sin2x^2
=4/sin2x^2-2sin2x^2/sin2x^2
=4/sin2x^2-2
=4/(2sinx*cosx)^2-2
=4/4sinˆ2x*cosˆ2x-2
=1/(sinˆ2xcosˆ2x)-2=左边
所以原式得证!