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Derivatives of the Trigonometric Functions

April 1st, 2009 |

Author: Merrin

In the big picture of differentiation, it is important to differentiate each type of basic elementary function. All of the basic elementary functions have derivatives in terms of the other elementary functions. For the case of the trigonometric functions their derivatives are specifically expressible in terms of trigonometric functions. The derivatives of more complicated elementary functions are done from these basic building blocks. In this section, we will derive the following derivatives of the trigonometric functions

$\cfrac{d}{dx} \sin x \qquad \cfrac{d}{dx} \cos x$
$\cfrac{d}{dx} \tan x \qquad \cfrac{d}{dx} \cot x$
$\cfrac{d}{dx} \sec x \qquad \cfrac{d}{dx} \csc x$

Knowing the derivatives of sine and cosine are necessary to derive the other derivatives. The easiest way to derive these derivatives is by using Euler’s formula.

$e^{ix} = \cos x + i \sin x$

We know that the derivativative of an exponential function is by the chain rule given by

$\cfrac{d}{dx} e^u = e^u \cfrac{du}{dx}$

We apply this rule to Euler’s formula and find

$\cfrac{d}{dx}e^{ix} = ie^{ix} = i(\cos x + i \sin x) =-\sin x +i\cos x$

But from the original formula we have

$\cfrac{d}{dx} e^{ix} = \cfrac{d}{dx} \cos x + i \cfrac{d}{dx} \sin x$

Matching up the real and imaginary parts we find that

$\cfrac{d}{dx} \sin x = +\cos x$
$\cfrac{d}{dx} \cos x = -\sin x$

Euler’s formula can be used a second way to compute the derivatives of sine and cosine. First write a negative Euler’s formula and add it to a positive Euler’s formula

$e^{+ix} = \cos x + i \sin x$
$e^{-ix} = \cos x - i \sin x$

If you add subtract the two formulas you find that

$\sin x = \cfrac{e^{ix} + e^{-ix}}{2}$

A similar formula for cosine can be derived by adding the two equations

$\cos x = \cfrac{e^{ix} + e^{-ix}}{2}$

These two formulas can be directly differentiated to give the other expression up to a plus or minus sign.

$\cfrac{d}{dx} \sin x = \cfrac{1}{2i}(e^{ix} -e^{-ix})'$
$(\sin x)' = \cfrac{1}{2}(e^{ix} + e^{-ix})$
$(\sin x)' = \cos x$

A similar procedure can be done for cosine to rederive the same result as above.

Assuming now that we know the derivatives for sine and cosine we can derive the other derivatives because they are just straightforward applications of the quotient rule or the reciprocal rule.

$\cfrac{d} {dx} \sec x = \cfrac{d} {dx} (\cos x)^{-1}$
$\cfrac{d}{dx} \csc x = \cfrac{d}{dx} (\sin x)^{-1}$
$\cfrac{d}{dx} \tan x = \cfrac{d}{dx} \cfrac{ \sin x} {\cos x} \qquad \cfrac{d}{dx} \cot x = \cfrac{d}{dx} \cfrac{ \cos x} {\sin x}$

Differentiating the first two can be done with the reciprocal rule. For the derivative of secant we have

$(\sec x)' = (-1)(\cos x)^{-2}(\cos x)'=\cfrac{\sin x}{\cos^2 x}$
$(\sec x)' = +\sec x \tan x$

For the derivative of cosecant we have

$(\csc x)' = (-1)(\sin x)^{-2}(\sin x)'= -\cfrac{\cos x}{\sin^2 x}$
$(\csc x)' = -\csc x \cot x$

Differentating tangent and cotangent can be done with the quotient rule. For the derivative of tangent we have

$(\tan x)' = \cfrac{(\sin x)'(\cos x) - (\sin x)(\cos x)'}{\cos^2 x}$
$(\tan x)' = \cfrac{\cos^2 x +\sin^2 x}{\cos^2 x}$
$(\tan x)' = \sec^2 x$

For the derivative of cotangent we have

$(\cot x)' = \cfrac{(\cos x)'(\sin x) - (\cos x)(\sin x)'}{\sin^2 x}$
$(\cot x)' = \cfrac{-\cos^2 x -\sin^2 x}{\sin^2 x}$
$(\cot x)' = -\csc^2 x$

For completeness it is worth mentioning a more standard method of calculating the derivatives of sine and cosine. Of course this method is to apply the definition directly. The basic method is to expand the trigonometric sum of two angles, factor the pieces and apply the trigonometric limits we derived by the squeeze theorem. These limits are

$\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\sin (\Delta x)}{\Delta x} =1 \qquad\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\cos (\Delta x)-1}{\Delta x} =0$

Here is how the derivative by the definition looks like for sine.

$\begin{array}{l}{\cfrac{d}{dx} \sin x} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\sin (x+\Delta x)-\sin x}{\Delta x} } \\ \\ {\cfrac{d}{dx} \sin x} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\sin x\cos (\Delta x)+\cos x\sin (\Delta x)-\sin x}{\Delta x}} \\ \\{\cfrac{d}{dx} \sin x} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\sin x(\cos (\Delta x)-1)+\cos x\sin (\Delta x)}{\Delta x} }\\ \\{\cfrac{d}{dx} \sin x} = {\sin x\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\cos (\Delta x)-1}{\Delta x} +\cos x\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\sin (\Delta x)}{\Delta x} }\\ \\{\cfrac{d}{dx} \sin x} = {\cos x} \end{array}$

Here is how the derivative by the definition looks like for cosine.

$\begin{array}{l}{\cfrac{d}{dx} \cos x} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\cos (x+\Delta x)-\cos x}{\Delta x} } \\ \\ {\cfrac{d}{dx} \cos x}={\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\cos x\cos (\Delta x)-\sin x\sin (\Delta x)-\cos x}{\Delta x} } \\ \\{\cfrac{d}{dx} \cos x} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\cos (x)(\cos (\Delta x)-1)-\sin (x)\sin (\Delta x)}{\Delta x} } \\ \\{\cfrac{d}{dx} \cos x} = {\cos x\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\cos (\Delta x)-1}{\Delta x} -\sin x\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\sin (\Delta x)}{\Delta x} }\\ \\{\cfrac{d}{dx} \cos x}={-\sin x} \end{array}$