请问这道题怎么求(a sin〖π/5〗+b cos〖π/5〗)/(a cos〖π/5〗-b sin〖π/5〗 )
请问这道题怎么求
(a sin〖π/5〗+b cos〖π/5〗)/(a cos〖π/5〗-b sin〖π/5〗 )=tan〖8/15 π〗,已知正实数a,b满足这一条件,求b/a的值
就这题
(a sin〖π/5〗+b cos〖π/5〗)/(a cos〖π/5〗-b sin〖π/5〗 )=tan〖8/15 π〗,已知正实数a,b满足这一条件,求b/a的值
就这题
赞 feiyeyue0713 幼苗
共回答了24个问题采纳率：95.8% 举报
tan(8π/15)=tan(π/3+π/5)=(tanπ/3+tanπ/5)/(1-tanπ/3*tanπ/5)
(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=(atanπ/5+b)/(a-btanπ/5)
=(b/a+tanπ/5)/(1-(b/a)tanπ/5)
所以b/a=tanπ/3=根号下3
(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=(atanπ/5+b)/(a-btanπ/5)
=(b/a+tanπ/5)/(1-(b/a)tanπ/5)
所以b/a=tanπ/3=根号下3
1年前 追问
4
哪里没看懂
简写的 你看两边都一模一样 所以b/a=tanπ/3 但这样答不行！你必须变为 (b/a+tanπ/5)/(1-(b/a)tanπ/5)=( tanπ/3+tanπ/5)/(1-tanπ/3*tanπ/5) 然后再变成二次函数（把tanπ/5看做未知数（只有一解）） 然后利用根判断定理求出b/a
可能相似的问题
-
1年前2个回答
-
请问sin（x+y）cos y-cos（x+y）sin y得多少
1年前1个回答
-
1年前1个回答
-
请问数学题,设sin a+cos a=3/5,则sin 2a=?
1年前4个回答
-
1年前1个回答
你能帮帮他们吗
精彩回答