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)