(asinπ/5 +bcosπ/5)/(acosπ/5 -bsinπ/5)=tan8π/15,则b/a等于 A.√3 B.√3/3 C.-√3 D.-√3/3(asinπ/5 +bcosπ/5)/(acosπ/5 -bsinπ/5)=tan8π/15,则b/a等于A.√3 B.√3/3 C.-√3 D.-√3/3

f(x)=asin(πx+a)+bcos(πx+b)+4 f(2010)=5 f(2013)= 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 知非零常数a,b满足asinπ/5+bcosπ/5／acosπ/5-bsinπ/5=tanπ/15,求b/a. 已知f(x)=aSin(πx+α)+bCos(πx+β)+4,若f(2009)=5,求f(2010)的值在线等.要过程.谢谢 f(x)=asin(πx+a)+bcos(πx+b) 且f(2006)=5 求f(2007)的值 f（x）=asin（πx+α）+bcos（πx+ β）,若 f（2003）=5,则f（2004）= f(x)=asin(πx+a)+bcos(πx+b) 且f(2006)=5 求f(2007)的值 设f(x)=asin(πx+Q)+bcos(πx+B)+4,且f(2003)=5,则f(2004)＝ 已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值 asinπ/4-bcosπ/4=√2/2(a-b)为什么asinπ/4-bcosπ/4=√2/2(a-b) f(2000)=asin(2000π+α)+bcos(2000π+β) =asinα+bcosβ 这一步怎么得出来的 在三角形ABC中,asin(π/2-A)+bcos(π-B)=0,则为 三角形? 已知非零常数 a,b满足（acosπ/5-bsinπ/5）/（asinπ/5+bcosπ/5）=1/（tan8π/15）,求b/a. 已知非零常数 a,b满足（acosπ/5-bsinπ/5）/（asinπ/5+bcosπ/5）=1/（tan8π/15）,求b/a. 详细过程 若f(x)=asin(πx+α)+bcos(πx+β)+4（a,b,α,β为非零实数）,f（0）=5,则f(2013)=?为什么=3? 设函数f(x)=asin(πx+α)+bcos(πx+β)(其中a,b,α,β为非零实数),若f(2006)=5,求f（2007）的值. 已知函数f（x）=asin（πx+α）+bcos（πx+β）+4,其中a,b,α,β都是非零实数,又知f（2008）=5,则f（2009）= 设函数f(x)=asin(π x+a)+bcos(π x+β)+4,其中a,b.a.β都是非零实数,若f(2011)=5,求f(2012)的值加4去掉