函数y=tan[(1+x/(a+b)]的周期为π,y=sin((a-b)x+π/4)的周期为3π,求a,b的值.

题目

函数y=tan[(1+x/(a+b)]的周期为π,y=sin((a-b)x+π/4)的周期为3π,求a,b的值.
答案a=5/6,b=1/6 a=1/6,b=5/6 a=-1/6,b=-5/6 a=-5/6,b=-1/6

答案

y=tan[(1+x/(a+b)]的周期T=π/[|1/(a+b)|]y=sin((a-b)x+π/4)的周期T=2π/(|a-b|)由题意,得π/[1/(a+b)]=π2π/(a-b)=3π∴a+b=1a-b=2/3∴a=5/6b=1/6 刚明白为什么是四种结果第二种情况是：π/[-1/(a+b)]=π2π/[-(a...