derivatives of trigonometric functions

Derivatives of trigonometric functions

tl;dr

  • \[(\sin\theta)^{\prime}=\cos\theta\]
  • \[(\cos\theta)^{\prime}=-\sin\theta\]
  • \[(\tan\theta)^{\prime}=1+\tan^{2}\theta=\sec^{2}\theta\]
  • \[(\csc\theta)^{\prime}=-\csc\theta\cot\theta\]
  • \[(\sec\theta)^{\prime}=\sec\theta\tan\theta\]
  • \[(\cot\theta)^{\prime}=-1-\cot^{2}\theta=-\csc^{2}\theta\]

\[\sin\theta\]

\begin{align*} \frac{d}{d\theta}\sin\theta&=\lim_{\delta\theta\to0}\frac{\sin(\theta+\delta\theta)-\sin\theta}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\frac{\sin\theta\cos\delta\theta+\cos\theta\sin\delta\theta-\sin\theta}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\frac{\cos\theta\sin\delta}{\delta\theta}+\frac{\sin\theta\cos\delta\theta-\sin\theta}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\frac{\sin\delta\theta}{\delta\theta}\cos\theta+\frac{\cos\delta\theta-1}{\delta\theta}\sin\theta\\ &=(1)(\cos\theta)+0\\ &=\cos\theta\\ \end{align*}

Proof for \[\lim_{\delta\theta\to0}\frac{\sin\delta\theta}{\delta\theta}=1\]

20240630-023446.png
This is a unit circle with radius one. Therefore it is known that \[BC=\sin\alpha\]. \[\triangle ABE=\frac{1}{2}(1)(\sin\alpha)=\frac{\sin\alpha}{2}\]. To find the area of the large triangle \[ADE\], we must find length \[DE\]. \[\tan\alpha=\frac{DE}{1}\], \[DE=\tan\alpha\]. \[\triangle ADE=\frac{1}{2}(1)(\tan\alpha)=\frac{\tan\alpha}{2}\]. Finally, the sector \[ABE\] has an area of \[\frac{1}{2}(1)^{2}(\alpha)=\frac{\alpha}{2}\].
Now that we have all the areas:

\begin{align*} \frac{\sin\alpha}{2}&<\frac{\alpha}{2}<\frac{\tan\alpha}{2}\\ \end{align*}

We divide everything by \[\frac{\sin\alpha}{2}\] and take the reciprocal:

\begin{align*} 1&<\frac{\alpha}{\sin\alpha}<\frac{1}{\cos\alpha}\\ 1&>\frac{\sin\alpha}{\alpha}>\cos\alpha\\ 1&\ge\lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}\ge\lim_{\alpha\to0}\cos\alpha\\ 1&\ge\lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}\ge1\\ \therefore&\lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}=1\\ \end{align*}

This limit can also be proved (and has been proven) with the Squeeze theorem.

\[\cos\theta\]

\begin{align*} \frac{d}{d\theta}\cos\theta&=\lim_{\delta\theta\to0}\frac{\cos(\theta+\delta\theta)-\cos\theta}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\frac{\cos\cos\delta\theta-\sin\theta\sin\delta\theta-\cos\theta}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\frac{\cos\delta\theta-1}{\delta\theta}\cos\theta-\frac{\sin\delta\theta}{\delta\theta}\sin\theta\\ &=0-1\sin\theta\\ &=-\sin\theta \end{align*}

\[\tan\theta\]

\begin{align*} \frac{d}{d\theta}\tan\theta&=\lim_{\delta\theta\to0}\frac{\tan(\theta+\delta\theta)-\tan\theta}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\frac{\frac{\tan\theta+\tan\delta\theta}{1-\tan\theta\tan\delta\theta}-\tan\theta}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\frac{(\tan\theta+\tan\delta\theta)-\tan\theta(1-\tan\theta\tan\delta\theta)}{\delta\theta(1-\tan\theta\tan\delta\theta)}\\ &=\lim_{\delta\theta\to0}\frac{\tan\theta+\tan\delta\theta-\tan\theta+\tan^{2}\theta+\tan\delta\theta}{\delta\theta(1-\tan\theta\tan\delta\theta)}\\ &=\lim_{\delta\theta\to0}\frac{\tan\delta\theta}{\delta\theta}\cdot\lim_{\delta\theta\to0}\frac{1+\tan^{2}\theta}{1-\tan\theta\tan\delta\theta}\\ &=1\cdot \frac{1+\tan^{2}\theta}{1}\\ &=1+\tan^{2}\theta\\ &=1+\frac{\sin^{2}\theta}{\cos^{2}\theta}\\ &=\frac{1}{\cos^{2}\theta}\\ &=\sec^{2}\theta\\ \end{align*}

Proof for \[\lim_{\delta\theta\to0}\frac{\tan\delta\theta}{\delta\theta}=1\]

\begin{align*} \lim_{\delta\theta\to0}\frac{\tan\delta\theta}{\delta\theta}&=\lim_{\delta\theta\to0}\frac{\frac{\sin\delta\theta}{\cos\delta\theta}}{\delta\theta}\\ &=\lim_{\delta\theta\to0}\left( \frac{\sin\delta\theta}{\delta\theta}\cdot \frac{1}{\cos\delta\theta} \right)\\ &=1 \end{align*}

\[\csc \theta\]

\[\csc\theta=\frac{1}{\sin\theta}\], using the reciprocal rule:

\begin{align*} (\csc\theta)^{\prime}&=\left( \frac{1}{\sin\theta} \right)^{\prime}\\ &=-\frac{(\sin\theta)^{\prime}}{(\sin\theta)^{2}}\\ &=-\frac{\cos\theta}{\sin^{2}\theta}\\ &=-\frac{1}{\sin\theta}\cdot \frac{\cos\theta}{\sin\theta}\\ &=-\csc\theta\cot\theta\\ \end{align*}

\[\sec\theta\]

\[\sec\theta=\frac{1}{\cos\theta}\], with reciprocal rule:

\begin{align*} (\sec\theta)^{\prime}&=\left( \frac{1}{\cos\theta} \right)^{\prime}\\ &=-\frac{(\cos\theta)^{\prime}}{\cos^{2}\theta}\\ &=\frac{\sin\theta}{\cos^{2}\theta}\\ &=\frac{1}{\cos\theta}\cdot \frac{\sin\theta}{\cos\theta}\\ &=\sec\theta\tan\theta\\ \end{align*}

\[\cot\theta\]

\[\cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}\], using the quotient rule:

\begin{align*} \left( \frac{\cos\theta}{\sin\theta} \right)^{\prime}&=\frac{(\cos\theta)^{\prime}\sin\theta-\cos\theta(\sin\theta)^{\prime}}{(\sin^{2}\theta)}\\ &=\frac{-\sin^{2}\theta-\cos^{2}\theta}{\sin^{2}\theta}\\ &=-\frac{\sin^{2}\theta}{\sin^{2}\theta}-\frac{\cos^{2}\theta}{\sin^{2}\theta}\\ &=-1-\cot^{2}\theta\\ \\ &=-\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin^{2}\theta}\\ &=-\frac{1}{\sin^{2}\theta}\\ &=-\csc^{2}\theta\\ \end{align*}
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