已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a. 详细过程
人气:334 ℃ 时间:2020-05-28 01:26:27
解答
(acosπ/5-bsinπ/5)/(asinπ/5 bcosπ/5)=1/(tan8π/15),
左边上下除以a.设b/a=k
cos8π/15sinπ/5+kcos8π/15cosπ/5=sin8π/15cosπ/5-ksinπ/5cos8π/15
kcos(8π/15-π/5)=sin(8π/15-π/5)
k=-tan(π/3)=-√3
b/a=-√3
推荐
- 已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a.
- 已知非零函数a.b满足:(asin(π/5)+bcos(π/5))/(acos(π/5)-bsin(π/5))=tan8π/5,求b/a的值
- 已知asin(γ+α)=bsin(γ+β),求证tanγ=bsinβ-asinα/acosα-bcosβ
- 已知非零实数a,b满足(asin(∏/5)+bcos(∏/5))/(acos(∏/5)-bsin(∏/5))=tan(8∏/15),求b/a的值
- 化简(Acosθ+Bsinθ)^2+(Asinθ-Bcosθ)^2
- -Hello,may I speak to Mrs Zhang,please?-Sorry,she is not in .She ___the school gym.
- 我最感动的时刻 - 作文 500字
- 一个数的2又5分之1倍是1又5分之4,这个数是多少?
猜你喜欢