已知f(x)=a*sin2x+b*tanx+1且f(-2)=4那么f(2+π)=

问题描述:

已知f(x)=a*sin2x+b*tanx+1且f(-2)=4那么f(2+π)=

-2

f(-2)=a*sin[2*(-2)] +b*tan(-2)+1=a*sin(-4) -b*tan2+1=-a*sin4-b*tan2+1根据f(-2)=4,可得出a*sin4+b*tan2=-3f(2+π)=a*sin[2*(2+π)] +b*tan(2+π)+1=a*sin(4+2π) +b*tan2+1=a*sin4+b*tan2+1 ①将①式代入,可求...