证明下列恒等式:(1)(cosx-1)²+sin²x=2-2cosx(2)1+tan²x=tanx/(sinxcosx)(3)sin²x+sin²xcos²x+cos²x=1
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![证明下列恒等式:(1)(cosx-1)²+sin²x=2-2cosx(2)1+tan²x=tanx/(sinxcosx)(3)sin²x+sin²xcos²x+cos²x=1](/uploads/image/z/8714026-10-6.jpg?t=%E8%AF%81%E6%98%8E%E4%B8%8B%E5%88%97%E6%81%92%E7%AD%89%E5%BC%8F%EF%BC%9A%281%29%28cosx-1%29%26%23178%3B%2Bsin%26%23178%3Bx%3D2-2cosx%282%291%2Btan%26%23178%3Bx%3Dtanx%2F%28sinxcosx%29%283%29sin%26%23178%3Bx%2Bsin%26%23178%3Bxcos%26%23178%3Bx%2Bcos%26%23178%3Bx%3D1)
证明下列恒等式:(1)(cosx-1)²+sin²x=2-2cosx(2)1+tan²x=tanx/(sinxcosx)(3)sin²x+sin²xcos²x+cos²x=1
证明下列恒等式:(1)(cosx-1)²+sin²x=2-2cosx
(2)1+tan²x=tanx/(sinxcosx)
(3)sin²x+sin²xcos²x+cos²x=1
证明下列恒等式:(1)(cosx-1)²+sin²x=2-2cosx(2)1+tan²x=tanx/(sinxcosx)(3)sin²x+sin²xcos²x+cos²x=1
证:
(1)(cosx-1)²+sin²x
=cos²x-2cosx+1+sin²x
=(cos²x+sin²x)+1-2cosx
=2-2cosx
(2)1+tan²x
=1+sin²x/cos²x
=(cos²x+sin²x)/cos²x
=1/cos²x
=sinx/(cos²xsinx)
=(sinx/cosx)/(sinxcosx)
=tanx/(sinxcosx)
(3)sin²x+sin²xcos²x+cos²x
=(sin²x+cos²x)+sin²xcos²x
=1+sin²xcos²x
【这题题目有抄错?】
:(1)(cosx-1)²+sin²x=2-2cosx展开sin^2x+xos^2x=1答案出来了(sin^2x)表sinx的平方
(3)是错识的
(2)也是可以详细点吗1:(cosx-1)²=cos^2x-2cosx+1代入 (cosx-1)²+sin²x=2-2cosx 3:sin²x+sin²xcos&...
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:(1)(cosx-1)²+sin²x=2-2cosx展开sin^2x+xos^2x=1答案出来了(sin^2x)表sinx的平方
(3)是错识的
(2)也是
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