化简:(1)−sin(π+α)+sin(−α)−tan(2π+α)tan(α−π)+cos(−α)+cos(π−α)
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化简:
(1)
(1)
−sin(π+α)+sin(−α)−tan(2π+α) |
tan(α−π)+cos(−α)+cos(π−α) |
![化简:(1)−sin(π+α)+sin(−α)−tan(2π+α)tan(α−π)+cos(−α)+cos(π−α)](/uploads/image/z/16673055-15-5.jpg?t=%E5%8C%96%E7%AE%80%EF%BC%9A%EF%BC%881%EF%BC%89%E2%88%92sin%28%CF%80%2B%CE%B1%29%2Bsin%28%E2%88%92%CE%B1%29%E2%88%92tan%282%CF%80%2B%CE%B1%29tan%28%CE%B1%E2%88%92%CF%80%29%2Bcos%28%E2%88%92%CE%B1%29%2Bcos%28%CF%80%E2%88%92%CE%B1%29)
(1)原式=
sinα−sinα−tanα
tanα+cosα−cosα=
−tanα
tanα=-1;
(2)当n为偶数时,原式=
sinα+sinα
sinαcosα=
2
cosα;
当n为奇数时,原式=
−sinα−sinα
−sinα•(−cosα)=-
2
cosα.
sinα−sinα−tanα
tanα+cosα−cosα=
−tanα
tanα=-1;
(2)当n为偶数时,原式=
sinα+sinα
sinαcosα=
2
cosα;
当n为奇数时,原式=
−sinα−sinα
−sinα•(−cosα)=-
2
cosα.
证明:1+sinα−cosα1+sinα+cosα=tanα2
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