求一道关于三角恒等变换的数学题
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/04 18:22:47
求一道关于三角恒等变换的数学题
![](http://img.wesiedu.com/upload/7/07/707a0a3287bef8ef22c56a3874d77ffa.jpg)
![](http://img.wesiedu.com/upload/7/07/707a0a3287bef8ef22c56a3874d77ffa.jpg)
![求一道关于三角恒等变换的数学题](/uploads/image/z/17935608-48-8.jpg?t=%E6%B1%82%E4%B8%80%E9%81%93%E5%85%B3%E4%BA%8E%E4%B8%89%E8%A7%92%E6%81%92%E7%AD%89%E5%8F%98%E6%8D%A2%E7%9A%84%E6%95%B0%E5%AD%A6%E9%A2%98)
由sin(Θ+kπ)=-2cos(Θ+kπ)得:
-2=sin(Θ+kπ)/cos(Θ+kπ)=tan(Θ+kπ)
由于tanΘ的周期是π,所以有:
tanΘ=tan(Θ+kπ)=-2.
题(1):[4sinΘ-2cosΘ]/[5cosΘ+3sinΘ]
上式的分子和分母同除cosΘ得:
[4tanΘ-2]/[5+3tanΘ]=(-8-2)/(5-6)=10
题(2):
(1/4)sin²Θ+(2/5)cos²Θ=(1/4)(1-cos²Θ)+(2/5)cos²Θ
=(1/4)-(1/4)cos²Θ+(2/5)cos²Θ=(1/4)+(3/20)cos²Θ
=(1/4)+(3/20)(1/sec²Θ)=(1/4)+(3/20)[1/(1+tan²Θ)]
=(1/4)+(3/20)[1/(1+4)]=(1/4)+(3/20)(1/5)=(1/4)+(3/100)
=7/25
-2=sin(Θ+kπ)/cos(Θ+kπ)=tan(Θ+kπ)
由于tanΘ的周期是π,所以有:
tanΘ=tan(Θ+kπ)=-2.
题(1):[4sinΘ-2cosΘ]/[5cosΘ+3sinΘ]
上式的分子和分母同除cosΘ得:
[4tanΘ-2]/[5+3tanΘ]=(-8-2)/(5-6)=10
题(2):
(1/4)sin²Θ+(2/5)cos²Θ=(1/4)(1-cos²Θ)+(2/5)cos²Θ
=(1/4)-(1/4)cos²Θ+(2/5)cos²Θ=(1/4)+(3/20)cos²Θ
=(1/4)+(3/20)(1/sec²Θ)=(1/4)+(3/20)[1/(1+tan²Θ)]
=(1/4)+(3/20)[1/(1+4)]=(1/4)+(3/20)(1/5)=(1/4)+(3/100)
=7/25