View source for Differentiation of trigonometric functions

{{short description|Mathematical process of finding the derivative of a trigonometric function}}
{{Trigonometry}}
{|class="wikitable" style="background-color:#FFFFFF; margin:1em 0 1em 1em" align="right" hspace="3" 
! Function
! Derivative
|-
| <math>\sin(x)</math>
| <math>\cos(x)</math>
|-
|<math>\cos(x)</math>
|<math>-\sin(x)</math>
|-
|<math>\tan(x)</math>
|<math>\sec^2(x)</math>
|-
|<math>\cot(x)</math>
|<math>-\csc^2(x)</math>
|-
|<math>\sec(x)</math>
|<math>\sec(x)\tan(x)</math>
|-
|<math>\csc(x)</math>
|<math>-\csc(x)\cot(x)</math>
|-
|<math>\arcsin(x)</math>
|<math>\frac{1}{\sqrt{1-x^2}}</math>
|-
|<math>\arccos(x)</math>
|<math>-\frac{1}{\sqrt{1-x^2}}</math>
|-
|<math>\arctan(x)</math>
|<math>\frac{1}{x^2+1}</math>
|-
|<math>\arccot(x)</math>
|<math>-\frac{1}{x^2+1}</math>
|-
|<math>\arcsec(x)</math>
|<math>\frac{1}{|x|\sqrt{x^2-1}}</math>
|-
|<math>\arccsc(x)</math>
|<math>-\frac{1}{|x|\sqrt{x^2-1}}</math>
|}
The '''differentiation of trigonometric functions''' is the mathematical process of finding the [[Derivative|derivative]] of a trigonometric function, or its rate of change with respect to a variable. For example, the derivative of the sine function is written sin′(''a'') = cos(''a''), meaning that the rate of change of sin(''x'') at a particular angle ''x = a'' is given by the cosine of that angle.

All derivatives of circular trigonometric functions can be found from those of sin(''x'') and cos(''x'') by means of the [[Quotient rule|quotient rule]] applied to functions such as tan(''x'') = sin(''x'')/cos(''x''). Knowing these derivatives, the derivatives of the [[Inverse trigonometric functions|inverse trigonometric functions]] are found using implicit differentiation.

==Proofs of derivatives of trigonometric functions==

===Limit of sin(θ)/θ as θ tends to 0===
[[File:limit circle FbN.jpeg|thumb|{{center|Circle, centre ''O'', radius 1}}]]

The diagram at right shows a circle with centre ''O'' and radius ''r ='' 1. Let two radii ''OA'' and ''OB'' make an arc  of θ radians. Since we are considering the limit as ''θ'' tends to zero, we may assume ''θ'' is a small positive number, say 0 < θ < ½ π in the first quadrant.

In the diagram, let ''R''<sub>1</sub> be the triangle ''OAB'', ''R''<sub>2</sub> the [[Circular sector|circular sector]] ''OAB'', and ''R''<sub>3</sub> the triangle ''OAC''. 

The [[Area of a triangle#Using trigonometry|area of triangle]] ''OAB'' is:

:<math> \mathrm{Area}(R_1
)
=\tfrac{1}{2} \ |OA| \ |OB|  \sin\theta = \tfrac{1}{2}\sin\theta \, . </math>

The [[Circular sector#Area|area of the circular sector]] ''OAB'' is:

:<math>\mathrm{Area}(R_2) 
=\tfrac{1}{2}\theta \, . </math>

The area of the triangle ''OAC'' is given by:

:<math> \mathrm{Area}(R_3
)
=\tfrac{1}{2} \ |OA| \ |AC| = \tfrac{1}{2} \tan\theta \, . </math>

Since each region is contained in the next, one has:

:<math>\text{Area}(R_1) < \text{Area}(R_2) < \text{Area}(R_3) \implies 
\tfrac{1}{2}\sin\theta < \tfrac{1}{2}\theta < \tfrac{1}{2}\tan\theta \, . </math>

Moreover, since {{nowrap|1=sin ''θ'' > 0}} in the first quadrant, we may divide through by ½ {{nowrap|sin ''θ''}}, giving:
:<math>1 < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta} \implies 1 > \frac{\sin\theta}{\theta} > \cos\theta \, . </math>

In the last step we took the reciprocals of the three positive terms, reversing the inequities.

[[File:Squeeze FbN.png|thumb|{{center|Squeeze: The curves {{nowrap|1=''y'' = 1}} and {{nowrap|1=''y'' = cos ''θ''}} shown in red, the curve {{nowrap|1=''y'' = sin(''θ'')/''θ''}} shown in blue.}}]]

We conclude that for 0 < θ < ½ π, the quantity {{nowrap|sin(''θ'')/''θ''}} is ''always'' less than 1 and ''always'' greater than cos(θ). Thus, as ''θ'' gets closer to 0, {{nowrap|sin(''θ'')/''θ''}} is "[[Squeeze theorem|squeezed]]" between a ceiling at height 1 and a floor at height {{nowrap|cos ''θ''}}, which rises towards 1; hence sin(''θ'')/''θ'' must tend to 1 as ''θ'' tends to 0 from the positive side:<blockquote><math>\lim_{\theta \to 0^+} \frac{\sin\theta}{\theta} = 1 \, . </math></blockquote>For the case where ''θ'' is a small negative number –½ π < θ < 0, we use the fact that sine is an odd function:

:<math>\lim_{\theta \to 0^-}\! \frac{\sin\theta}{\theta} 
\ =\ 
\lim_{\theta\to 0^+}\!\frac{\sin(-\theta)}{-\theta} 
\ =\ 
\lim_{\theta \to 0^+}\!\frac{-\sin\theta}{-\theta} 
\ =\ 
\lim_{\theta\to 0^+}\!\frac{\sin\theta}{\theta} \ =\ 
 1 \, . </math>

===Limit of (cos(θ)-1)/θ as θ tends to 0===
The last section enables us to calculate this new limit relatively easily. This is done by employing a simple trick. In this calculation, the sign of ''θ'' is unimportant. 
:<math> \lim_{\theta \to 0}\, \frac{\cos\theta - 1}{\theta} 
\ =\ 
\lim_{\theta \to 0} \left( \frac{\cos\theta - 1}{\theta} \right)\!\! \left( \frac{\cos\theta + 1}{\cos\theta + 1} \right) 
\ =\ 
\lim_{\theta \to 0}\, \frac{\cos^2\!\theta - 1}{\theta\,(\cos\theta + 1)} . </math>

Using {{nowrap|1=cos<sup>2</sup>''θ'' – 1 = –sin<sup>2</sup>''θ'',}}
the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:

:<math> \lim_{\theta \to 0}\,\frac{\cos\theta - 1}{\theta}
\ =\ 
\lim_{\theta \to 0}\, \frac{-\sin^2\theta}{\theta(\cos\theta+1)} 
\ =\ 
\left( -\lim_{\theta \to 0} \frac{\sin\theta}{\theta}\right)\! \left( \lim_{\theta \to 0}\,\frac{\sin\theta}{\cos\theta + 1} \right) 
\ =\ 
(-1)\left(\frac{0}{2}\right) = 0 \, . </math>

===Limit of tan(θ)/θ as θ tends to 0===
Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:
:<math>
  \lim_{\theta\to 0} \frac{\tan\theta}{\theta}
  \ =\ 
 \left(\lim_{\theta\to 0} \frac{\sin\theta}{\theta}\right)\!
 \left( \lim_{\theta\to 0} \frac{1}{\cos\theta}\right)
 \ =\ 
 (1)(1)
 \ =\ 
  1 \, . </math>

===Derivative of the sine function===
We calculate the derivative of the sine function from the [[Derivative#Definition via difference quotients|limit definition]]:
:<math> \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\sin\theta = \lim_{\delta \to 0}  \frac{\sin(\theta + \delta) - \sin \theta}{\delta} . </math>

Using the [[List of trigonometric identities#Angle sum and difference identities|angle addition formula]] {{nowrap|1=sin(α+β) = sin α cos β + sin β cos α}}, we have:
:<math> \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\sin\theta 
 =
\lim_{\delta \to 0} \frac{\sin\theta\cos\delta + \sin\delta\cos\theta-\sin\theta}{\delta} 
 =
\lim_{\delta \to 0} \left( \frac{\sin\delta}{\delta} \cos\theta 
+ \frac{\cos\delta -1}{\delta}\sin\theta \right) . </math>
Using the limits for the sine and cosine functions:
:<math> \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\sin\theta 
 =
(1)\cos\theta + (0)\sin\theta 
 =
\cos\theta \, . </math>

===Derivative of the cosine function===

====From the definition of derivative====
We again calculate the derivative of the cosine function from the limit definition:

:<math> \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\cos\theta 
 =
\lim_{\delta \to 0} \frac{\cos(\theta+\delta)-\cos\theta}{\delta} . </math>

Using the angle addition formula {{nowrap|1=cos(α+β) = cos α cos β – sin α sin β}}, we have:
:<math> \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\cos\theta 
 =
\lim_{\delta \to 0} \frac{\cos\theta\cos\delta - \sin\theta\sin\delta-\cos\theta}{\delta} 
 =
\lim_{\delta \to 0} \left(\frac{\cos\delta -1}{\delta}\cos\theta \,-\, \frac{\sin\delta}{\delta} \sin\theta \right) . </math>
Using the limits for the sine and cosine functions:
:<math> \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\cos\theta 
 = (0) \cos\theta - (1) \sin\theta = -\sin\theta \, . </math>

====From the chain rule====
To compute the derivative of the cosine function from the chain rule, first observe the following three facts:
:<math>\cos\theta = \sin\left(\tfrac{\pi}{2}-\theta\right)</math>
:<math>\sin\theta = \cos\left(\tfrac{\pi}{2}-\theta\right)</math>
:<math>\tfrac{\operatorname{d}}{\operatorname{d}\!\theta} \sin\theta = \cos\theta</math>
The first and the second are [[List of trigonometric identities#Symmetry|trigonometric identities]], and the third is proven above. Using these three facts, we can write the following,
:<math>\tfrac{\operatorname{d}}{\operatorname{d}\!\theta} \cos\theta = \tfrac{\operatorname{d}}{\operatorname{d}\!\theta} \sin\left(\tfrac{\pi}{2}-\theta\right)</math>
We can differentiate this using the [[Chain rule|chain rule]]. Letting <math>f(x) = \sin x,\ \ g(\theta) =\tfrac{\pi}{2}-\theta</math>, we have:

:<math>\tfrac{\operatorname{d}}{\operatorname{d}\!\theta} f\!\left(g\!\left(\theta\right)\right) = f^\prime\!\left(g\!\left(\theta\right)\right) \cdot g^\prime\!\left(\theta\right) = \cos\left(\tfrac{\pi}{2}-\theta\right) \cdot (0-1) = -\sin\theta</math>.

Therefore, we have proven that
:<math>\tfrac{\operatorname{d}}{\operatorname{d}\!\theta} \cos\theta = -\sin\theta</math>.

===Derivative of the tangent function===

====From the definition of derivative====
To calculate the derivative of the tangent function tan ''&theta;'', we use [[Derivative#Definition via difference quotients|first principles]]. By definition:
:<math>
 \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\tan\theta
 = \lim_{\delta \to 0} \left( \frac{\tan(\theta+\delta)-\tan\theta}{\delta} \right) .
</math>
Using the well-known angle formula {{nowrap|1=tan(α+β) = (tan α + tan β) / (1 - tan α tan β)}}, we have:
:<math>
 \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\tan\theta
 = \lim_{\delta \to 0} \left[ \frac{\frac{\tan\theta + \tan\delta}{1 - \tan\theta\tan\delta} - \tan\theta}{\delta} \right]
 = \lim_{\delta \to 0} \left[ \frac{\tan\theta + \tan\delta - \tan\theta + \tan^2\theta\tan\delta}{\delta \left( 1 - \tan\theta\tan\delta \right)} \right] .
</math>
Using the fact that the limit of a product is the product of the limits:
:<math>
 \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\tan\theta
 = \lim_{\delta \to 0} \frac{\tan\delta}{\delta} \times \lim_{\delta \to 0} \left( \frac{1 + \tan^2\theta}{1 - \tan\theta\tan\delta} \right) .
</math>
Using the limit for the tangent function, and the fact that tan ''&delta;'' tends to 0 as &delta; tends to 0:
:<math>
 \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\tan\theta
 = 1 \times \frac{1 + \tan^2\theta}{1 - 0} = 1 + \tan^2\theta .
</math>
We see immediately that:
:<math>
 \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\tan\theta
 = 1 + \frac{\sin^2\theta}{\cos^2\theta}
 = \frac{\cos^2\theta + \sin^2\theta}{\cos^2\theta}
 = \frac{1}{\cos^2\theta}
 = \sec^2\theta \, .
</math>

====From the quotient rule====
One can also compute the derivative of the tangent function using the [[Quotient rule|quotient rule]].
:<math>\frac{\operatorname{d}}{\operatorname{d}\!\theta} \tan\theta 
 = \frac{\operatorname{d}}{\operatorname{d}\!\theta} \frac{\sin\theta}{\cos\theta}
 = \frac{\left(\sin\theta\right)^\prime \cdot \cos\theta - \sin\theta \cdot \left(\cos\theta\right)^\prime}{ \cos^2 \theta }
 = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}
</math>
The numerator can be simplified to 1 by the [[List of trigonometric identities#Pythagorean identities|Pythagorean identity]], giving us,
:<math>\frac{1}{\cos^2 \theta} = \sec^2 \theta</math>
Therefore,
:<math>\frac{\operatorname{d}}{\operatorname{d}\!\theta} \tan\theta = \sec^2 \theta</math>

==Proofs of derivatives of inverse trigonometric functions==
The following derivatives are found by setting a [[Variable (mathematics)|variable]] ''y'' equal to the inverse trigonometric function that we wish to take the derivative of. Using implicit differentiation and then solving for ''dy''/''dx'', the derivative of the inverse function is found in terms of ''y''. To convert ''dy''/''dx'' back into being in terms of ''x'', we can draw a reference triangle on the unit circle, letting ''θ'' be y. Using the [[Pythagorean theorem]] and the definition of the regular trigonometric functions, we can finally express ''dy''/''dx'' in terms of ''x''.

===Differentiating the inverse sine function===
We let

:<math>y=\arcsin x\,\!</math>

Where

:<math>-\frac{\pi}{2}\le y \le \frac{\pi}{2}</math>

Then

:<math>\sin y=x\,\!</math>

Taking the derivative with respect to <math>x</math> on both sides and solving for dy/dx:

:<math>{d \over dx}\sin y={d \over dx}x</math>

:<math>\cos y \cdot {dy \over dx} = 1\,\!</math>

Substituting <math> \cos y = \sqrt{1-\sin^2 y}</math> in from above,

:<math>\sqrt{1-\sin^2 y} \cdot {dy \over dx} =1</math>

Substituting <math>x=\sin y</math> in from above,

:<math>\sqrt{1-x^2} \cdot {dy \over dx} =1</math>

:<math>{dy \over dx}=\frac{1}{\sqrt{1-x^2}}</math>

===Differentiating the inverse cosine function===
We let

:<math>y=\arccos x\,\!</math>

Where

:<math>0 \le y \le \pi</math>

Then

:<math>\cos y=x\,\!</math>

Taking the derivative with respect to <math>x</math> on both sides and solving for dy/dx:

:<math>{d \over dx}\cos y={d \over dx}x</math>

:<math>-\sin y \cdot {dy \over dx} =1</math>

Substituting <math>\sin y = \sqrt{1-\cos^2 y}\,\!</math> in from above, we get

:<math>-\sqrt{1-\cos^2 y} \cdot {dy \over dx} =1</math>

Substituting <math>x=\cos y\,\!</math> in from above, we get

:<math>-\sqrt{1-x^2} \cdot {dy \over dx} =1</math>

:<math>{dy \over dx} = -\frac{1}{\sqrt{1-x^2}}</math>

Alternatively, once the derivative of <math>\arcsin x</math> is established, the derivative of <math>\arccos x</math> follows immediately by differentiating the identity <math>\arcsin x+\arccos x=\pi/2</math> so that <math>(\arccos x)'=-(\arcsin x)'</math>.

===Differentiating the inverse tangent function===

We let

:<math>y=\arctan x\,\!</math>

Where

:<math>-\frac{\pi}{2} < y < \frac{\pi}{2}</math>

Then

:<math>\tan y=x\,\!</math>

Taking the derivative with respect to <math>x</math> on both sides and solving for dy/dx:

:<math>{d \over dx}\tan y={d \over dx}x</math>

Left side:

:<math>
 {d \over dx}\tan y
 = \sec^2 y \cdot {dy \over dx}
 = (1 + \tan^2 y) {dy \over dx}
</math> using the Pythagorean identity

Right side:

:<math>{d \over dx}x = 1</math>

Therefore,

:<math>(1+\tan^2 y){dy \over dx}=1</math>

Substituting <math>x=\tan y\,\!</math> in from above, we get

:<math>(1+x^2){dy \over dx}=1</math>

:<math>{dy \over dx}=\frac{1}{1+x^2}</math>

===Differentiating the inverse cotangent function===

We let

:<math>y=\arccot x</math>

where <math>0<y<\pi</math>. Then

:<math>\cot y=x</math>

Taking the derivative with respect to <math>x</math> on both sides and solving for dy/dx:

:<math>\frac{d}{dx}\cot y=\frac{d}{dx}x</math>

Left side:

:<math>
 {d \over dx}\cot y
 = -\csc^2 y \cdot {dy \over dx}
 = -(1 + \cot^2 y) {dy \over dx}
</math> using the Pythagorean identity

Right side:

:<math>{d \over dx}x = 1</math>

Therefore,

:<math>-(1+\cot^2y)\frac{dy}{dx}=1</math>

Substituting <math>x=\cot y</math>,

:<math>-(1+x^2)\frac{dy}{dx}=1</math>

:<math>\frac{dy}{dx}=-\frac{1}{1+x^2}</math>
Alternatively, as the derivative of <math>\arctan x</math> is derived as shown above, then using the identity <math>\arctan x+\arccot x=\dfrac{\pi}{2}</math> follows immediately that<math display="block">\begin{align}
\dfrac{d}{dx}\arccot x 
&=\dfrac{d}{dx}\left(\dfrac{\pi}{2}-\arctan x\right)\\
&=-\dfrac{1}{1+x^2}
\end{align}</math>

===Differentiating the inverse secant function===
====Using implicit differentiation====
Let

:<math> y = \arcsec x\ \mid |x| \geq 1</math>

Then

:<math> x = \sec y \mid \ y \in \left [0,\frac{\pi}{2} \right )\cup \left (\frac{\pi}{2},\pi \right]
</math>

:<math> \frac{dx}{dy} = \sec y \tan y = |x|\sqrt{x^2-1}</math>
(The absolute value in the expression is necessary as the product of secant and tangent in the interval of y is always nonnegative, while the radical <math>\sqrt{x^2-1}</math> is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)

:<math> \frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}}</math>

====Using the chain rule====
Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using the [[Chain rule|chain rule]].

Let

:<math> y = \arcsec x = \arccos \left(\frac{1}{x}\right) </math>

Where

:<math> |x| \geq 1 </math> and <math> y \in \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right] </math>

Then, applying the chain rule to <math> \arccos \left(\frac{1}{x}\right) </math>:

:<math> \frac{dy}{dx} = -\frac{1}{\sqrt{1-(\frac{1}{x})^2}} \cdot \left(-\frac{1}{x^2}\right)
 = \frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}
 = \frac{1}{x^2\frac{\sqrt{x^2-1}}{\sqrt{x^2}}}
 = \frac{1}{\sqrt{x^2}\sqrt{x^2-1}}
 = \frac{1}{|x|\sqrt{x^2-1}} </math>

===Differentiating the inverse cosecant function===
====Using implicit differentiation====
Let

:<math>y = \arccsc x\ \mid |x| \geq 1</math>

Then

:<math> x = \csc y\ \mid \ y \in \left [-\frac{\pi}{2},0 \right )\cup \left (0,\frac{\pi}{2} \right]</math>

:<math> \frac{dx}{dy} = -\csc y \cot y = -|x|\sqrt{x^2-1}</math>
(The absolute value in the expression is necessary as the product of cosecant and cotangent in the interval of y is always nonnegative, while the radical <math>\sqrt{x^2-1}</math> is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)
:<math> \frac{dy}{dx} = \frac{-1}{|x|\sqrt{x^2-1}}</math>

====Using the chain rule====
Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using the [[Chain rule|chain rule]].

Let

:<math> y = \arccsc x = \arcsin \left(\frac{1}{x}\right) </math>

Where

:<math> |x| \geq 1 </math> and <math> y \in \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] </math>

Then, applying the chain rule to <math> \arcsin \left(\frac{1}{x}\right) </math>:

:<math> \frac{dy}{dx} =\frac{1}{\sqrt{1-(\frac{1}{x})^2}} \cdot \left(-\frac{1}{x^2}\right)
 = -\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}
 = -\frac{1}{x^2\frac{\sqrt{x^2-1}}{\sqrt{x^2}}}
 = -\frac{1}{\sqrt{x^2}\sqrt{x^2-1}}
 = -\frac{1}{|x|\sqrt{x^2-1}} </math>

==See also==

* {{annotated link|Calculus}}
* {{annotated link|Derivative}}
* {{annotated link|Differentiation rules}}
* {{annotated link|General Leibniz rule}}
* {{annotated link|Inverse functions and differentiation}}
* {{annotated link|Linearity of differentiation}}
* {{annotated link|List of integrals of inverse trigonometric functions}}
* {{annotated link|List of trigonometric identities}}
* {{annotated link|Trigonometry}}

==References==
{{Reflist}}

==Bibliography==

*''[[Abramowitz and Stegun|Handbook of Mathematical Functions]]'', Edited by Abramowitz and Stegun, National Bureau of Standards, Applied Mathematics Series, 55 (1964)

{{DEFAULTSORT:Differentiation Of Trigonometric Functions}}

[[Category:Articles containing proofs]]
[[Category:Differential calculus]]
[[Category:Differential equations]]
[[Category:Mathematical identities]]

{{Sourceattribution|Differentiation of trigonometric functions}}