假设函数 f f f 充分光滑，即 f f f 在 x 0 x_0 x0​ 点处任意阶导数存在，取一小邻域 U ( x 0 ) \small U(x_0) U(x0​)，则 ∀   x ∈ U ( x 0 ) \small \forall\,x\in U(x_0) ∀x∈U(x0​)， f ( x ) \small f(x) f(x) 都可以展开为泰勒级数，即 f ( x ) = ∑ n = 0 ∞ f ( n ) ( x 0 ) n ! ( x − x 0 ) n = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ f(x)=\sum_{n=0}^{\infin}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots f(x)=n=0∑∞​n!f(n)(x0​)​(x−x0​)n=f(x0​)+f′(x0​)(x−x0​)+2!f′′(x0​)​(x−x0​)2+⋯考虑三角函数 sin ⁡ x , cos ⁡ x \sin x,\cos x sinx,cosx 的任意阶导函数， sin ⁡ ′ ( x ) = cos ⁡ ( x ) = sin ⁡ ( x + π 2 ) sin ⁡ ′ ′ ( x ) = cos ⁡ ′ ( x ) = sin ⁡ ′ ( x + π 2 ) = cos ⁡ ( x + π 2 ) = sin ⁡ ( x + π 2 + π 2 ) = sin ⁡ ( x + π ) \begin{aligned}\sin'(x)&=\cos(x)=\sin (x+\frac{\pi}{2})\\ \sin''(x)&=\cos'(x)=\sin'(x+\frac{\pi}{2})\\&=\cos (x+\frac{\pi}{2})=\sin (x+\frac{\pi}{2}+\frac{\pi}{2})\\&=\sin (x+\pi) \end{aligned} sin′(x)sin′′(x)​=cos(x)=sin(x+2π​)=cos′(x)=sin′(x+2π​)=cos(x+2π​)=sin(x+2π​+2π​)=sin(x+π)​假设 sin ⁡ ( n ) ( x ) = sin ⁡ ( x + n π / 2 ) \sin^{(n)}(x)=\sin(x+n\pi/2) sin(n)(x)=sin(x+nπ/2)，下面我们用 数学归纳法 来验证一下.

  首先，当 n = 0 n=0 n=0 时， sin ⁡ ( 0 ) ( x ) = sin ⁡ ( x ) \sin^{(0)}(x)=\sin(x) sin(0)(x)=sin(x)，结论显然成立.

  其次，假设当 n = k ( k ≥ 0 ) n=k(k\geq 0) n=k(k≥0) 时结论成立，即 sin ⁡ ( k ) ( x ) = sin ⁡ ( x + k π / 2 ) \displaystyle\sin^{(k)}(x)=\sin(x+k\pi/2) sin(k)(x)=sin(x+kπ/2).

  考虑 n = k + 1 n=k+1 n=k+1 的情形，此时 sin ⁡ ( k + 1 ) ( x ) = d ( sin ⁡ ( k ) ( x ) ) d x = d ( sin ⁡ ( x + k π / 2 ) ) d x = cos ⁡ ( x + k π / 2 ) = sin ⁡ ( x + k π / 2 + π / 2 ) = sin ⁡ ( x + ( k + 1 ) π / 2 ) \begin{aligned} \sin^{(k+1)}(x)&=\frac{d(\sin^{(k)}(x))}{dx}\\&=\frac{d(\sin(x+k\pi/2))}{dx}\\&=\cos(x+k\pi/2)\\&=\sin(x+k\pi/2+\pi/2)\\&=\sin(x+(k+1)\pi/2) \end{aligned} sin(k+1)(x)​=dxd(sin(k)(x))​=dxd(sin(x+kπ/2))​=cos(x+kπ/2)=sin(x+kπ/2+π/2)=sin(x+(k+1)π/2)​  所以，当 n = k + 1 n=k+1 n=k+1 时，结论照样成立.

  由归纳法原理， ∀   n ∈ N ,   sin ⁡ ( n ) ( x ) = sin ⁡ ( x + n π / 2 ) \forall\,n\in N,\,\sin^{(n)}(x)=\sin(x+n\pi/2) ∀n∈N,sin(n)(x)=sin(x+nπ/2).

则 cos ⁡ ( n ) ( x ) = sin ⁡ ( n + 1 ) ( x ) = sin ⁡ ( x + n π / 2 + π / 2 ) = cos ⁡ ( x + n π / 2 ) \cos^{(n)}(x)=\sin^{(n+1)}(x)=\sin(x+n\pi/2+\pi/2)=\cos(x+n\pi/2) cos(n)(x)=sin(n+1)(x)=sin(x+nπ/2+π/2)=cos(x+nπ/2).

  特别地，取 x = 0 x=0 x=0，

sin ⁡ ( 2 k ) ( 0 ) = sin ⁡ ( k π ) = 0 sin ⁡ ( 2 k + 1 ) ( 0 ) = sin ⁡ ( k π + π / 2 ) = ( − 1 ) k cos ⁡ ( 2 k ) ( 0 ) = cos ⁡ ( k π ) = ( − 1 ) k cos ⁡ ( 2 k + 1 ) ( 0 ) = cos ⁡ ( k π + π / 2 ) = 0 ( k = 0 , 1 , 2 , ⋯   ) \begin{aligned} &\sin^{(2k)}(0)=\sin(k\pi)=0 \\ &\sin^{(2k+1)}(0)=\sin(k\pi+\pi/2)=(-1)^{k} \\ \\ &\cos^{(2k)}(0)=\cos(k\pi)=(-1)^{k} \\ &\cos^{(2k+1)}(0)=\cos(k\pi+\pi/2)=0 \\ &(k=0,1,2,\cdots) \end{aligned} ​sin(2k)(0)=sin(kπ)