f(x)=asin(πx+a)+bcos(πx+b),且f(2009)=3,求f(2010)
问题描述:
f(x)=asin(πx+a)+bcos(πx+b),且f(2009)=3,求f(2010)
答
f(2009)=asin(2009π+a)+bcos(2009π+b)
f(2010)=asin(2010π+a)+bcos(2010π+b)
=asin(π+2009π+a)+bcos(π+2009π+b)
=-[asin(2009π+a)+bcos(2009π+b)]
=-f(2009)
=-3