asint-bsin3t

用最小公倍数法：

asint：最小正周期2π

bsin3t：最小正周期2π/3

∴最小正周期是2π

f(t)=asint-bsin3t

f(t+2π)=aasin(t+2π)-bsin3(t+2π)

=asin(t+2π)-bsin(3t+6π)

=asint-bsin3t

=f(t)

asint-bsin3t用最小公倍数法：asint：最小正周期2πbsin3t：最小正周期2π/3∴最小正周期是2πf(t)=asint-bsin3tf(t+2π)=aasin(t+2π)-bsin3(t+2π)=asin(t+2π)-bsin(3t+6π)=asint-bsin3t=f(t)