高一二倍角三角函数1.求证:(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/22.化简:(1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x)3.已知sin(π/4 -x)=5/13,0
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![高一二倍角三角函数1.求证:(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/22.化简:(1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x)3.已知sin(π/4 -x)=5/13,0](/uploads/image/z/14834972-20-2.jpg?t=%E9%AB%98%E4%B8%80%E4%BA%8C%E5%80%8D%E8%A7%92%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B01.%E6%B1%82%E8%AF%81%EF%BC%9A%28sinx%2Bcosx-1%29%28sinx-cosx%2B1%29%2Fsin2x%3Dtanx%2F22.%E5%8C%96%E7%AE%80%3A%EF%BC%881%2Bsin4x-cos4x%29%2F%281%2Bsin4x%2Bcos4x%29+-+%281%2Bsin4x%2Bcos4x%29%2F%281%2Bsin4x-cos4x%293.%E5%B7%B2%E7%9F%A5sin%28%CF%80%2F4+-x%29%3D5%2F13%2C0)
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