[函数] totutu三角函数n阶导数

\begin{align*}
(\sin ax)^{(0)}&=\sin ax,\\
(\sin ax)^{(1)}&=a\cos ax=a\sin\left(ax+\frac\pi2\right),\\
(\sin ax)^{(2)}&=-a^2\sin ax=a^2\sin(ax+\pi),\\
(\sin ax)^{(3)}&=-a^3\cos ax=a^3\sin\left(ax+\frac{3\pi}2\right),\\
(\sin ax)^{(4)}&=a^4\sin ax=a^2\sin(ax+2\pi),\\
\ldots&\ldots
\end{align*}
归纳出
\begin{equation}
(\sin ax)^{(n)}=a^n\sin\left(ax+\frac{n\pi}2\right).
\end{equation}
用数学归纳法证之，假设 $n=k$ 成立那么 $n=k+1$ 时
\begin{align*}
(\sin ax)^{(k+1)}&=\left(a^k\sin\left(ax+\frac{k\pi}2\right)\right)'\\
&=a^k\cdot a\cos\left(ax+\frac{k\pi}2\right)\\
&=a^{k+1}\sin\left(ax+\frac{k\pi}2+\frac\pi2\right)\\
&=a^{k+1}\sin\left(ax+\frac{(k+1)\pi}2\right),
\end{align*}
即得证。由此亦可得 cos 的
\[(\cos ax)^{(n)}=\left(\sin\left(ax+\frac\pi2\right)\right)^{(n)}
=a^n\sin\left(ax+\frac\pi2+\frac{n\pi}2\right)
=a^n\cos\left(ax+\frac{n\pi}2\right).\]

计算 $(\sin^2x)^{(n)}$

\[(\sin^2x)^{(n)}=\left(\frac{1-\cos2x}2\right)^{(n)}=-\frac12(\cos2x)^{(n)}
=-\frac12\cdot2^n\cos\left(2x+\frac{n\pi}2\right)=-2^{n-1}\cos\left(2x+\frac{n\pi}2\right).\]
或

\begin{align*}
(\sin^2x)^{(n)}&=(\sin x\cdot\sin x)^{(n)}\\
&=\sum_{k=0}^n C_n^k(\sin x)^{(k)}(\sin x)^{(n-k)}\\
&=\sum_{k=0}^n C_n^k\sin\left(x+\frac{k\pi}2\right)\sin\left(x+\frac{(n-k)\pi}2\right)\\
&=\sum_{k=0}^n \frac{C_n^k}2\left(\cos\left(\frac{k\pi}2-\frac{(n-k)\pi}2\right)
-\cos\left(2x+\frac{n\pi}2\right)\right)\\
&=\sum_{k=0}^n \frac{C_n^k}2\left((-1)^k\cos\frac{n\pi}2-\cos\left(2x+\frac{n\pi}2\right)\right)\\
&=\frac{\cos\frac{n\pi}2}2\sum_{k=0}^n C_n^k(-1)^k
- \frac{\cos\left(2x+\frac{n\pi}2\right)}2\sum_{k=0}^n C_n^k\\
&=-\frac{\cos\left(2x+\frac{n\pi}2\right)}2\cdot2^n\\
&=-2^{n-1}\cos\left(2x+\frac{n\pi}2\right).
\end{align*}