Since 2 π is the period of the sinusoidal function y = sin x , x ∈ R, since 2\pi is the period of the sinusoidal function y=\sin x,x\in R,Since 2 π is a sine function y=sinx,x∈For the period of R , we only need to prove that any positive number less than 2 π is not the period of that function. It is only necessary to prove that any positive number less than 2\pi is not a period of that function.Now it is only necessary to prove that any positive number less than 2 π is not a period of that function.
Counter-evidence: Counter-evidence:Counter-evidence:
Let T be the minimum positive period of y = sin x , x ∈ R, and 0 < T < 2 π . Then, according to the definition of a periodic function, when x is any value, sin ( x + T ) = sinx Let x = π / 2 and substitute it into the above formula to get: sin ( π / 2 + T ) = sin ( π / 2 ) = 1 , that is, cos T = 1 , which is the same as when T ∈ ( 0 , 2 π ), cos T < 1 is a contradiction, so the assumption does not hold. Therefore, the minimum positive period is 2 π . Let T be the minimum positive period of y=\sin x,x\in R, and 0<T<2π.\\Then, according to the definition of periodic function, when x is any value, There is sin(x+T) = sin x\\Let x=π/2, substitute into the above formula to get:\\sin(π/2+T) = sin(π/2) = 1,\\namely cos T = 1,\\This contradicts with T∈(0,2π), cos T < 1, so the assumption is not valid. \\Therefore the minimum positive period is 2\pi.Let T be y=sinx,x∈The minimum positive period of R , and 0<T<2 p.m. _Then, according to the definition of periodic function, when x is any value, there is s in ( x+T)=s in xmake x=π /2 , substitute into the above formula to get:s in ( π /2+T)=s in ( π /2 )=1,That is cos T=1,This is related to T∈(0,2 π ) , cos T<1 contradiction ,Therefore the assumption does not hold.So the minimum positive period is 2 π .
For the function y = A sin ( ω x + φ ) + k , for the function y=Asin(ωx+φ)+k,for function y=Asin(ωx+f )+k , where A , ω , φ , k are all constants, we can deduce its minimum positive period through the following steps. Among them, A, ω, φ, and k are all constants, and we can derive its minimum positive period through the following steps.where A ,oh _f ,k is a constant, we can deduce its minimum positive period through the following steps.
任 x 0 、 x 1 ∈ R ， 有 sin ( x 0 + 2 π ) = sin ( x 0 )、 ω x 1 + φ = x 0 ， 则 sin ( ω x 1 + φ ) = sin ( x 0 )、 sin [ ω ( x 1 + T ) + φ ] = sin ( x 0 + 2 π ) 所什 ω T = 2 π 任x_{0}、x_{1}\in R，\\有sin(x_{0} +2\pi) = sin (x_{0})、ωx_{1}+φ=x_{0}，\\则sin(ωx_{1}+φ)=sin (x_{0})、sin[ω (x_{1}+T)+φ]=sin(x_{0}+2\pi)\\soωT=2\pilet x0、x1∈R，There are s in ( x0+2 p )=sin(x0)、ωx1+Phi=x0，Then s in ( ω x1+f )=sin(x0) 、s in [ ω ( x1+T)+f ]=sin(x0+2 p )So ω T=2 p.m
Or sinx satisfies f ( x + T 0 ) = − f ( x ) , or sin x satisfies f(x+T_{0}) =- f (x),Or s in x satisfies f ( x+T0)=− f ( x ) , that is, sin ( x + T 0 ) = − sinx so sin ( x + 2 T 0 ) = − sin ( x + T 0 ) and therefore sin ( x + 2 T 0 ) = sinx because sin ( x + π ) = − sinx , then the minimum positive period is 2 π, namely sin(x+T_{0}) =- sin x\\so sin(x+2T_{0}) =-sin(x+T_{0 }) , so sin(x+2T_{0}) =sin x\\Because sin(x+π) =- sin x, the minimum positive period is 2\piThat is, s in ( x+T0)=- s in xSo s in ( x+2T _0)=−sin(x+T0) , so s in ( x+2T _0)=s in xBecause s in ( x+p )=− s in x , then the minimum positive period is 2 π
同理：任 x 0 、 x 1 ∈ R , 有 sin ( x 0 + π ) = − sin ( x 0 ) 、 ω x 1 + φ = x 0 , 即 sin ( ω x 1 + φ + π ) = − sin ( ω x 1 + φ ) sin [ ( ω x 1 + φ ) + π + π ] = − sin ( ω x 1 + φ + π ) 同理：任x_{0}、x_{1}\in R ，\\有sin(x_{0}+\pi) = -sin (x_{0})、ωx_{1}+φ=x_{0}，\\即sin(ωx_{1}+φ+\pi )=-sin(ωx_{1}+φ)\\sin[(ωx_{1}+φ)+\pi+\pi]=-sin(ωx_{1}+φ+\pi)Similarly: let x0、x1∈R，There are s in ( x0+p )=−sin(x0)、ωx1+Phi=x0，That is, s in ( ω x1+Phi+p )=− s in ( ω x1+f )s in [( ω x1+f )+Pi+p ]=− s in ( ω x1+Phi+p )
sin [ ω ( x + t ) + φ ] = − sin ( ω x + φ ) sin [ ω ( x + 2 t ) + φ ] = − sin [ ω ( x + t ) + φ ] sin[ω(x +t)+φ]=-sin(ωx+φ)\\sin[ω(x+2t)+φ]=-sin[ω(x+t)+φ]s in [ ω ( x+t)+f ]=− s in ( ω x+f )s in [ ω ( x+2 t )+f ]=− s in [ ω ( x+t)+f ]
而 sin [ ω ( x + 2 t ) + φ ] = sin [ ( ω x + φ ) + 2 t ω ] ，则 2 t ω = 2 π 而sin[ω(x+2t)+φ]=sin[ (ωx+φ)+2tω]，则2tω=2\pi而s in [ ω ( x+2 t )+f ]=s in [( ω x+f )+2 t ω ] , then 2 t ω=2 p.m