> 数学 >
求证:
cosα
1+sinα
sinα
1+cosα
2(cosα−sinα)
1+sinα+cosα
人气:359 ℃ 时间:2020-03-25 09:43:31
解答
证明:左边=
1+sina+cosa
1+sina+cosa
(
cosa
1+sina
sina
1+cosa
)

=
1
1+sina+cosa
[
(1+sina+cosa)cosa
1+sina
(1+cosa+sina)sina
1+cosa
]

=
1
1+sina+cosa
[cosa+
cos2a
1+sina
−sina−
sin2a
1+cosa
]

=
1
1+sina+cosa
(cosa+1−sina−sina−1+cosa)

=
2(cosa−sina)
1+sina+cosa
=右边.
故原式成立.
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