sin[α-(π/3)]+cos[α+(13π/6)]化简还有sin[α-(5π/4)]+cos[α+(π/2)]
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 04:16:06
![sin[α-(π/3)]+cos[α+(13π/6)]化简还有sin[α-(5π/4)]+cos[α+(π/2)]](/uploads/image/z/14652939-3-9.jpg?t=sin%5B%CE%B1-%28%CF%80%2F3%29%5D%2Bcos%5B%CE%B1%2B%2813%CF%80%2F6%29%5D%E5%8C%96%E7%AE%80%E8%BF%98%E6%9C%89sin%5B%CE%B1-%285%CF%80%2F4%29%5D%2Bcos%5B%CE%B1%2B%28%CF%80%2F2%29%5D)
sin[α-(π/3)]+cos[α+(13π/6)]化简还有sin[α-(5π/4)]+cos[α+(π/2)]
sin[α-(π/3)]+cos[α+(13π/6)]化简
还有sin[α-(5π/4)]+cos[α+(π/2)]
sin[α-(π/3)]+cos[α+(13π/6)]化简还有sin[α-(5π/4)]+cos[α+(π/2)]
sin[α-(π/3)]+cos[α+(13π/6)]
=sin(α-π/3)+cos(α+π/6)
=sinαcosπ/3-cosαsinπ/3+cosαcosπ/6-sinαsinπ/6
=(1/2)sinα-(√3/2)cosα+(√3/2)cosα-(1/2)sinα
=0
sin[α-(5π/4)]+cos[α+(π/2)]
=sinαcos5π/4-cosαsin5π/4+cosαcosπ/2-sinαsinπ/2
=-√2/2sinα+√2/2cosα+0-sinα
=(-1-√2/2)sinα+√2/2cosα
(1)因为 cos[a+(13派/6)]=cos[a+(派/6)]
=cos[(派/2)+(a--派/3)]
=--sin[a--(派/3)]
...
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(1)因为 cos[a+(13派/6)]=cos[a+(派/6)]
=cos[(派/2)+(a--派/3)]
=--sin[a--(派/3)]
所以 sin[a--(派/3)]+cos[a+(13派/6)]
=sin[a--(派/3)]--sin[a--(派/3)]
=0.
(2) 因为 sin[a--(5派/4)]=sinacos(5派/4)--cosasin(5派/4)
=--sinacos(派/4)--cosasin(派/4)
=[--(根号2)/2](sina+cosa),
cos[a+(派/2)]=--sina
所以 sin[a--(5派/4)]+cos[a+(派/2)]
=[--1--(根号2)/2]sina--[(根号2)/2]cosa.
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