Theorem 6.1.1. Derivative of sine is cosine.
Let \(f(x) = \sin{x}\text{.}\) Then, \(f'(x) = \cos{x}\text{.}\) In other words,
\begin{equation*}
\boxed{\frac{d}{dx} \sin{x} = \cos{x}}
\end{equation*}
Section 6.1 Derivatives of Trigonometric Functions
Note that in calculus, the inputs of trigonometric functions are always measured in radians.
Subsection 6.1.1 Derivative of Sine
Consider the derivative of the sine function \(f(x) = \sin{x}\text{.}\) Since the sine functon is periodic, it is sufficient to consider it over the interval \([0,2\pi]\text{.}\)
First, the sine function has horizontal tangent lines at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\text{,}\) so \(f'(x) = 0\) there. At \(x = 0\text{,}\) the slope of the tangent is 1 (this can be intuitively confirmed graphically), so \(f'(0) = 1\text{.}\) In addition, at \(x = \pi\text{,}\) \(f'(\pi) = -1\text{.}\) From these points, it is intuitive that these points trace out the cosine function. Consider this Desmos applet, where you can see the slope of \(\sin{x}\) trace out the cosine curve 1 . In summary,
Subsection 6.1.2 Derivative of Cosine
Similarly, consider this Desmos applet, which shows the slope of the tangent lines of \(\cos{x}\) is \(-\sin{x}\) 2 .
Theorem 6.1.2. Derivative of cosine is negative of sine.
Let \(f(x) = \cos{x}\text{.}\) Then, \(f'(x) = -\sin{x}\text{.}\) In other words,
\begin{equation*}
\boxed{\frac{d}{dx} \cos{x} = -\sin{x}}
\end{equation*}
Subsection 6.1.3 Limit Definition of Derivative of Sine and Cosine
For the sine function, consider the derivative, using the limit definition,
\begin{equation*}
f'(x) = \lim_{h \to 0} \frac{\sin{(x + h)} - \sin{x}}{h}
\end{equation*}
The term \(\sin{(x + h)}\) can be expanded using the sine addition identity,
\begin{align*}
\amp = \lim_{h \to 0} \frac{\sin{x}\cos{h} + \cos{x}\sin{h} - \sin{x}}{h} \amp\amp \text{by the sine addition identity}\\
\amp = \lim_{h \to 0} \frac{\sin{x}(\cos{h} - 1) + \cos{x} \sin{h}}{h}
\end{align*}
Then, we can use the sum rule to split this into two limits,
\begin{equation*}
= \lim_{h \to 0} \frac{\sin{x}(\cos{h} - 1)}{h} + \lim_{h \to 0} \frac{\cos{x} \sin{h}}{h}
\end{equation*}
Then, factor out \(\sin{x}\) and \(\cos{x}\) respectively from each limit, as they are constants with respect to the limit \(h \to 0\text{,}\)
\begin{equation*}
= \sin{x} \lim_{h \to 0} \frac{\cos{h} - 1}{h} + \cos{x} \lim_{h \to 0} \frac{\sin{h}}{h}
\end{equation*}
Then, the derivative depends on the two limits,
\begin{equation*}
\lim_{h \to 0} \frac{\sin{h}}{h} \qquad \text{and} \qquad \lim_{h \to 0} \frac{\cos{h} - 1}{h}
\end{equation*}
which are both of indeterminate form, because substituting in \(h = 0\) gives \(\frac{0}{0}\text{.}\)
Subsection 6.1.4 Two Important Limits
Consider the limit,
\begin{equation*}
\lim_{h \to 0} \frac{\sin{h}}{h}
\end{equation*}
Consider this Desmos applet 3 . Here, the horizontal axis represents the \(h\)-axis. Notice that as \(h \to 0\text{,}\) the expression \(\frac{\sin{h}}{h}\) seems to approach 1. In summary,
Fact 6.1.3.
\begin{equation*}
\boxed{\lim_{x \to 0} \frac{\sin{x}}{x} = 1}
\end{equation*}
Proof.
A more precise proof can be done using the squeeze theorem. Note that the proof only works if \(x\) is in radians. If the angle is measured in degrees, then the area of a sector formula is different.
Similarly, consider the limit,
\begin{equation*}
\lim_{h \to 0} \frac{\cos{h} - 1}{h}
\end{equation*}
Consider this Desmos applet 4 , where you can see that as \(h \to 0\text{,}\) the expression approaches 0. In summary,
Fact 6.1.4.
\begin{equation*}
\boxed{\lim_{x \to 0} \frac{\cos{x} - 1}{x} = 0}
\end{equation*}
Proof.
There is a similar geometric interpretation, however it can be derived algebraically from the previous limit.
\begin{align*}
\lim_{x \to 0} \frac{\cos{x} - 1}{x} \amp = \lim_{x \to 0} \frac{\cos{x} - 1}{x} \cdot \frac{\cos{x} + 1}{\cos{x} + 1}\\
\amp = \lim_{x \to 0} \frac{\cos^2{x} - 1}{x(\cos{x} + 1)}\\
\amp = \lim_{x \to 0} \frac{\sin^2{x}}{x(\cos{x} + 1)}\\
\amp = \lim_{x \to 0} \underbrace{\frac{\sin{x}}{x}}_{\to 1} \cdot \underbrace{\frac{\sin{x}}{\cos{x} + 1}}_{\to \frac{0}{2} = 0}\\
\amp = 0
\end{align*}
This means that both of these expressions have a removable discontinuity at \(x = 0\text{.}\)
Subsection 6.1.5 Derivative of Sine and Cosine Conclusion
Recall from before,
\begin{equation*}
f'(x) = \sin{x} \lim_{h \to 0} \frac{\cos{h} - 1}{h} + \cos{x} \lim_{h \to 0} \frac{\sin{h}}{h}
\end{equation*}
Then, substituting the known values of the two limits,
\begin{align*}
f'(x) \amp = \sin{x} \cdot 0 + \cos{x} \cdot 1\\
f'(x) \amp = \cos{x}
\end{align*}
To prove the derivative of cosine, one method is to use the complementary angle identities to write cosine in terms of sine (and vise versa),
\begin{equation*}
\sin{\brac{\frac{\pi}{2} - x}} = \cos{x} \qquad \text{and} \qquad \cos{\brac{\frac{\pi}{2} - x}} = \sin{x}
\end{equation*}
Proof using the sine derivative, and the chain rule.
Let \(f(x) = \cos{x}\text{.}\) Then,
\begin{align*}
f'(x) \amp = \frac{d}{dx} \cos{x}\\
\amp = \frac{d}{dx} \sin{\brac{\frac{\pi}{2} - x}}\\
\amp = (-1) \cos{\brac{\frac{\pi}{2} - x}} \amp\amp \text{by the chain rule}\\
\amp = -\sin{x}
\end{align*}
On the other hand, the derivative of cosine can be shown directly from the limit definition, in a similar way as the the sine derivative.
Direct proof.
Let \(f(x) = \cos{x}\text{.}\) Using the cosine addition formula \(\cos{(x + h)} = \cos{x} \cos{h} - \sin{x} \sin{h}\text{,}\)
\begin{align*}
f'(x) \amp = \lim_{h \to 0} \frac{\cos{(x + h)} - \cos{x}}{h}\\
\amp = \lim_{h \to 0} \frac{\cos{x}\cos{h} - \sin{x}\sin{h} - \cos{x}}{h} \amp\amp \text{by the cosine addition identity}\\
\amp = \lim_{h \to 0} \frac{\cos{x}(\cos{h} - 1) - \sin{x}\sin{h}}{h}
\end{align*}
In a similar way, splitting up the limit and factoring out the constants \(\cos{x}\) and \(\sin{x}\text{,}\)
\begin{align*}
\amp = \cos{x} \lim_{h \to 0} \frac{\cos{h} - 1}{h} - \sin{x} \lim_{h \to 0} \frac{\sin{h}}{h}\\
\amp = \cos{x} \cdot 0 - \sin{x} \cdot 1 \amp\amp \text{by the limits from before}\\
\amp = -\sin{x}
\end{align*}
Subsection 6.1.6 Derivatives of Other Trigonometric Functions
The tangent function, and the three reciprocal functions (secant, cosecant, and cotangent) can all be expressed in terms of sine and cosine functions. In particular,
\begin{equation*}
\tan{x} = \frac{\sin{x}}{\cos{x}} \qquad \sec{x} = \frac{1}{\cos{x}} \qquad \csc{x} = \frac{1}{\sin{x}} \qquad \cot{x} = \frac{\cos{x}}{\sin{x}}
\end{equation*}
In this way, their derivatives follow from the derivatives of sine and cosine, along with the product and quotient rules.
Theorem 6.1.5. Derivative of tangent.
If \(f(x) = \tan{x}\text{,}\) then \(f'(x) = \sec^2{x}\text{.}\) In other words,
\begin{equation*}
\boxed{\frac{d}{dx} \tan{x} = \sec^2{x}}
\end{equation*}
Proof.
Since \(\tan{x} = \frac{\sin{x}}{\cos{x}}\text{,}\) by the quotient rule,
\begin{align*}
f'(x) \amp = \frac{d}{dx} \frac{\sin{x}}{\cos{x}}\\
\amp = \frac{\cos{x}(\cos{x}) - \sin{x}(-\sin{x})}{(\cos{x})^2}\\
\amp = \frac{\cos^2{x} + \sin^2{x}}{\cos^2{x}}\\
\amp = \frac{1}{\cos^2{x}} \amp\amp \text{by the Pythagorean identity}\\
\amp = \sec^2{x}
\end{align*}
Theorem 6.1.6. Derivative of secant.
If \(f(x) = \sec{x}\text{,}\) then \(f'(x) = \sec{x} \tan{x}\text{.}\) In other words,
\begin{equation*}
\boxed{\frac{d}{dx} \sec{x} = \sec{x} \tan{x}}
\end{equation*}
Proof.
Since \(\sec{x} = \frac{1}{\cos{x}}\text{,}\) then by the reciprocal rule,
\begin{align*}
f'(x) \amp = \frac{d}{dx} \frac{1}{\cos{x}}\\
\amp = -\frac{-\sin{x}}{(\cos{x})^2}\\
\amp = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}}\\
\amp = \sec{x} \tan{x}
\end{align*}
Theorem 6.1.7. Derivative of cosecant.
If \(f(x) = \csc{x}\text{,}\) then \(f'(x) = -\csc{x} \cot{x}\text{.}\) In other words,
\begin{equation*}
\boxed{\frac{d}{dx} \csc{x} = -\csc{x} \cot{x}}
\end{equation*}
Proof.
Since \(\csc{x} = \frac{1}{\sin{x}}\text{,}\) by the reciprocal rule,
\begin{align*}
f'(x) \amp = \frac{d}{dx} \frac{1}{\sin{x}}\\
\amp = -\frac{\cos{x}}{(\sin{x})^2}\\
\amp = -\frac{1}{\sin{x}} \cdot \frac{\cos{x}}{\sin{x}}\\
\amp = -\csc{x} \cot{x}
\end{align*}
Theorem 6.1.8. Derivative of cotangent.
If \(f(x) = \cot{x}\text{,}\) then \(f'(x) = -\csc^2{x}\text{.}\) In other words,
\begin{equation*}
\boxed{\frac{d}{dx} \cot{x} = -\csc^2{x}}
\end{equation*}
Proof.
Since \(\cot{x} = \frac{\cos{x}}{\sin{x}}\text{,}\) by the quotient rule,
\begin{align*}
f'(x) \amp = \frac{d}{dx} \frac{\cos{x}}{\sin{x}}\\
\amp = \frac{\sin{x}(-\sin{x}) - \cos{x}(\cos{x})}{(\sin{x})^2}\\
\amp = -\frac{\sin^2{x} + \cos^2{x}}{\sin^2{x}}\\
\amp = -\frac{1}{\sin^2{x}} \amp\amp \text{by the Pythagorean identity}\\
\amp = -\csc^2{x}
\end{align*}
Subsection 6.1.7 Summary of Trigonometric Derivatives
In summary,
Theorem 6.1.9. Trigonometric derivatives.
\begin{equation*}
\boxed{\begin{array}{ll}
\dfrac{d}{dx} \sin{x} = \cos{x} \amp \dfrac{d}{dx} \csc{x} = -\csc{x} \cot{x} \\[10pt]
\dfrac{d}{dx} \cos{x} = -\sin{x} \amp \dfrac{d}{dx} \sec{x} = \sec{x} \tan{x} \\[10pt]
\dfrac{d}{dx} \tan{x} = \sec^2{x} \amp \dfrac{d}{dx} \cot{x} = -\csc^2{x}
\end{array}}
\end{equation*}
Note that these “nice” formulas only apply if \(x\) is measured in radians. If \(x\) is measured in degrees, then there are extra factors of \(\frac{\pi}{180}\text{.}\) In fact, this is the main reason radians are taught in pre-calculus.
Subsection 6.1.8 Derivatives of Cofunctions
Notice that the derivatives of the “cofunctions” (cosine, cosecant, and cotangent) all contain negative signs. Further, notice that if we denote \(\operatorname{co}{f(x)}\) to be the “cofunction” of \(f(x)\text{,}\) then 6 trigonometric derivatives follow the rule,
\begin{equation*}
\frac{d}{dx} \operatorname{co}(f(x)) = - \operatorname{co} \brac{\frac{d}{dx} f(x)}
\end{equation*}
For example, th derivative of \(\sec{x}\) is \(\sec{x} \tan{x}\text{,}\) so to remember the derivative of \(\csc{x}\text{,}\) which is the cofunction of \(\sec{x}\text{,}\) switch the functions \(\sec{x} \tan{x}\) to their cofunctions \(\csc{x} \cot{x}\text{,}\) and put a negative sign, to get \(-\csc{x} \cot{x}\text{.}\)
In fact, it turns out it is also true that,
\begin{equation*}
\frac{d^2}{dx^2} \operatorname{co}(f(x)) = \operatorname{co}\brac{\frac{d^2}{dx^2} f(x)}
\end{equation*}
Subsection 6.1.9 Continuity of Sine and Cosine
It is intuitively clear that the sine and cosine functions \(\sin{x}\) and \(\cos{x}\) are continuous functions. However, this can be proved more rigorously, using a similar technique to that to prove the two limits above.
Assuming that \(\cos{x}\) and \(\sin{x}\) are continuous at \(x = 0\text{,}\) this can be used to prove that the functions are continuous on \(\mathbb{R}\text{.}\) That is,
\begin{equation*}
\lim_{h \to 0} \cos{h} = \cos{0} = 1 \qquad \text{and} \qquad \lim_{h \to 0} \sin{h} = \sin{0} = 0
\end{equation*}
Proof that sine is continuous.
\begin{align*}
\lim_{h \to 0} \sin{(x + h)} \amp = \lim_{h \to 0} \brac{\sin{x} \cos{h} - \cos{x} \sin{h}} \amp\amp \text{using the sine addition formula}\\
\amp = \sin{x} \lim_{h \to 0} \cos{h} - \cos{x} \lim_{h \to 0} \sin{h} \amp\amp \text{as $\cos{x}, \sin{x}$ are constant with respect to the limits}\\
\amp = \sin{x} \cdot 1 - \cos{x} \cdot 0 \amp\amp\\
\amp = \sin{x}
\end{align*}
Proof that cosine is continuous.
\begin{align*}
\lim_{h \to 0} \cos{(x + h)} \amp = \lim_{h \to 0} \brac{\cos{x} \cos{h} - \sin{x} \sin{h}} \amp\amp \text{using the cosine addition formula}\\
\amp = \cos{x} \lim_{h \to 0} \cos{h} - \sin{x} \lim_{h \to 0} \sin{h} \amp\amp \text{as $\cos{x}, \sin{x}$ are constant with respect to the limits}\\
\amp = \cos{x} \cdot 1 - \sin{x} \cdot 0 \amp\amp\\
\amp = \cos{x}
\end{align*}
Subsection 6.1.10 Cyclical Pattern of Sine Derivatives
Consider the first few derivatives of sine \(f(x) = \sin{x}\text{,}\)
\begin{align*}
f'(x) \amp = \cos{x}\\
f''(x) \amp = -\sin{x}\\
f'''(x) \amp = -\cos{x}\\
f^{(4)}(x) \amp = \sin{x}
\end{align*}
which returns back to \(\sin{x}\text{.}\) Thus, this pattern continues.
Theorem 6.1.10. Derivatives of sine.
\begin{equation*}
\frac{d^n}{dx^n} \sin{x} = \begin{cases} \sin{x} \amp n = 0, 4, 8, \dots \\
\cos{x} \amp n = 1, 5, 9, \dots \\
-\sin{x} \amp n = 2, 6, 10, \dots \\
-\cos{x} \amp n = 3, 7, 11, \dots \end{cases}
\end{equation*}
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