求证:(sin x+cos x−1)(sin x−cos x+1)sin&nbs
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求证:
=tan
(sin x+cos x−1)(sin x−cos x+1) |
sin 2x |
x |
2 |
![求证:(sin x+cos x−1)(sin x−cos x+1)sin&nbs](/uploads/image/z/613921-49-1.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9A%28sin%26nbsp%3Bx%2Bcos%26nbsp%3Bx%E2%88%921%29%28sin%26nbsp%3Bx%E2%88%92cos%26nbsp%3Bx%2B1%29sin%26nbs)
证明:左边=
[sin x−(1−cos x)](sin x+1−cos x)
sin 2x
=
(2sin
x
2cos
x
2−2sin2
x
2)(2sin
x
2cos
x
2+2sin2
x
2)
2sin xcosx
=
4sin 2
x
2(cos
x
2−sin
x
2)(cos
x
2+sin
x
2)
2sin xcosx
=
4sin 2
x
2(cos2
x
2−sin2
x
2)
4sin
x
2cos
x
2cosx
=
sin
x
2
cos
x
2
=tan
x
2=右边,
故等式得证.
[sin x−(1−cos x)](sin x+1−cos x)
sin 2x
=
(2sin
x
2cos
x
2−2sin2
x
2)(2sin
x
2cos
x
2+2sin2
x
2)
2sin xcosx
=
4sin 2
x
2(cos
x
2−sin
x
2)(cos
x
2+sin
x
2)
2sin xcosx
=
4sin 2
x
2(cos2
x
2−sin2
x
2)
4sin
x
2cos
x
2cosx
=
sin
x
2
cos
x
2
=tan
x
2=右边,
故等式得证.
求证:(sin x+cos x−1)(sin x−cos x+1)sin&nbs
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