Differentiation of trigonometric functions: Information from Answers.com

Trigonometry
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Reference
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Euclidean theory
Law of sines Law of cosines Law of tangents Pythagorean theorem
Calculus
Trigonometric substitution Integrals of functions Derivatives of functions Integrals of inverses

Function	Derivative
$sin(x)$	$cos(x)$
$cos(x)$	$- sin(x)$
$tan(x)$	$sec 2 (x)$
$cot(x)$	$- csc 2 (x)$
$sec(x)$	$sec(x)tan(x)$
$csc(x)$	$- csc(x)cot(x)$
$arcsin(x)$	$\frac{1}{\sqrt{1-x^2}}$
$arccos(x)$	$\frac{-1}{\sqrt{1-x^2}}$
$arctan(x)$	$\frac{1}{x^2+1}$

The differentiation of trigonometric functions is the mathematical process of finding the rate at which a trigonometric function changes with respect to a variable--the derivative of the trigonometric function. Commonplace trigonometric functions include sin(x), cos(x) and tan(x). For example, in differentiating f(x) = sin(x), one is calculating a function f ′(x) which computes the rate of change of sin(x) at a particular point a. The value of the rate of change at a is thus given by f ′(a). Knowledge of differentiation from first principles is required, along with competence in the use of trigonometric identities and limits. All functions involve the arbitrary variable x, with all differentiation performed with respect to x.

It turns out that once one knows the deriatives of sin(x) and cos(x), one can easily compute the derivatives of the other circular trigonometric functions because they can all be expressed in terms of sine or cosine; the quotient rule is then implemented to differentiate this expression. Proofs of the derivatives of sin(x) and cos(x) are given in the proofs section; the results are quoted in order to give proofs of the derivatives of the other circular trigonometric functions. Finding the derivatives of the inverse trigonometric functions involves using implicit differentiation and the derivatives of regular trigonometric functions also given in the proofs section.

1 Derivatives of trigonometric functions and their inverses
2 Proofs of derivative of the sine and cosine functions
3 Proofs of derivatives of inverse trigonometric functions
4 See also
5 References
6 Bibliography

Derivatives of trigonometric functions and their inverses

$\left(\sin(x)\right)' = \cos(x)$

$\left(\cos(x)\right)' = -\sin(x)$

$\left(\tan(x)\right)' = \left(\frac{\sin(x)}{\cos(x)}\right)' = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} = \sec^2(x)$

$\left(\cot(x)\right)' = \left(\frac{\cos(x)}{\sin(x)}\right)' = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)} = -(1+\cot^2(x)) = -\csc^2(x)$

$\left(\sec(x)\right)' = \left(\frac{1}{\cos(x)}\right)' = \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)}.\frac{\sin(x)}{\cos(x)} = \sec(x)\tan(x)$

$\left(\csc(x)\right)' = \left(\frac{1}{\sin(x)}\right)' = -\frac{\cos(x)}{\sin^2(x)} = -\frac{1}{\sin(x)}.\frac{\cos(x)}{\sin(x)} = -\csc(x)\cot(x)$

$\left(\arcsin(x)\right)' = \frac{1}{\sqrt{1-x^2}}$

$\left(\arccos(x)\right)' = \frac{-1}{\sqrt{1-x^2}}$

$\left(\arctan(x)\right)' = \frac{1}{x^2+1}$

Proofs of derivative of the sine and cosine functions

Proof of $\lim_{x\to 0}{\frac{\sin(x)}{x}}=1$ ^[1]

From the above graph, the area of triangle OPA < the area of sector OPA < the area of triangle OAQ.

Let the angle that is subtended by the arc AP to be x and the radius of the circle to be r,

When $0<x<\frac{\pi}{2}$

The area of the triangle OPA is $\frac{1}{2}r^2 \sin(x)$ . The area of the sector OPA is $\frac{1}{2}r^2 x$ . The area of Triangle OAQ is $\frac{1}{2}r^2\tan(x)$ . Then we have

sin(x) < x < tan(x)

$\frac{\sin(x)}{x}<1$

x cos(x) < sin(x)

$\frac{\sin(x)}{x}>\cos(x)$

$\cos(x)<\frac{\sin(x)}{x}<1$

When $-\frac{\pi}{2}<x<0,$

- sin(x) < - x < - tan(x)

sin(x) > x

$\frac{\sin(x)}{x}<1$

x > tan(x)

x cos(x) > s i n (x)

$\frac{\sin(x)}{x}>cos(x)$

$\cos(x)<\frac{\sin(x)}{x}<1$

Therefore, when $-\frac{\pi}{2}<x<\frac{\pi}{2}$ for $x \ne 0$

$\cos(x)<\frac{\sin(x)}{x}<1$

$\lim_{x\to 0}{\cos(x)=1}$

By Squeeze Theorem, we conclude that

$\lim_{x\to 0}{\frac{\sin(x)}{x}}=1$

Differentiating the sine function

Definition of a derivative of a function f(x):

$f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h}$

Therefore if f(x) = sin(x)

$f'(x)=\lim_{h\to 0}{\sin(x+h)-\sin(x)\over h}$

From the trigonometric identity

$\sin(A)-\sin(B)=2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$

we can say

$f'(x)=\lim_{h\to 0}{\frac{2\cos(x+\frac{h}{2})\sin(\frac{h}{2})}{h}}$

Since

$\lim_{x\to 0}{\frac{\sin(x)}{x}}=1$

Therefore

$f' (x)=(\lim_{h\to 0}{\cos(x+\frac{h}{2})})(\lim_{h\to 0}{\frac{\sin(\frac{h}{2})}{\frac{h}{2}}})=\cos(x)$

Differentiating the cosine function

From the trigonometric identity

$\cos(x)=\sin(\frac{\pi}{2}-x)$

As shown above, since

$\frac{d}{dx}(\sin(x))=\cos(x)$

Therefore,

$\frac{d}{dx}(\cos(x))=\frac{d}{dx}(\sin(\frac{\pi}{2}-x))=-\cos(\frac{\pi}{2}-x)=-\sin(x)$

Proofs of derivatives of inverse trigonometric functions

The following derivatives are found by setting a 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 theta 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

$y=\arcsin x\,\!$

Where

$-\frac{\pi}{2}\le y \le \frac{\pi}{2}$

Then

$\sin y=x\,\!$

Using implicit differentiation and solving for dy/dx:

${d \over dx}\sin y={d \over dx}x$

${dy \over dx}\cos y=1\,\!$

Substituting $\cos y = \sqrt{1-\sin^2 y}$ in from above,

${dy \over dx}\sqrt{1-\sin^2 y}=1$

Substituting $x = sin y$ in from above,

${dy \over dx}\sqrt{1-x^2}=1$

${dy \over dx}=\frac{1}{\sqrt{1-x^2}}$

Differentiating the inverse cosine function

We let

$y=\arccos x\,\!$

Where

$0 \le y \le \pi$

Then

$\cos y=x\,\!$

Using implicit differentiation and solving for dy/dx:

${d \over dx}\cos y={d \over dx}x$

$-{dy \over dx}\sin y=1$

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

$-{dy \over dx}\sqrt{1-\cos^2 y} =1$

Substituting $x=\cos y\,\!$ in from above, we get

$-{dy \over dx}\sqrt{1-x^2} =1$

${dy \over dx} = -\frac{1}{\sqrt{1-x^2}}$

Differentiating the inverse tangent function

We let

$y=\arctan x\,\!$

Where

$-\frac{\pi}{2} < y < \frac{\pi}{2}$

Then

$\tan y=x\,\!$

Using implicit differentiation and solving for dy/dx:

${d \over dx}\tan y={d \over dx}x$

${dy \over dx}\sec^2 y=1$

Substituting $1+\tan^2 y = \sec^2 y\,\!$ into the above,

${dy \over dx}(1+\tan^2 y)=1$

Substituting $x=\tan y\,\!$ in from above,

${dy \over dx}(1+x^2)=1$

${dy \over dx}=\frac{1}{x^2+1}$

It is simpler to derive the two members of the relation : $tan(arctan(x)) = x$

knowing that the derivative of tan(x) is $1 + tan 2 (x)$

References

^ The Squeeze Theorem Applied to Useful Trig Limits

Bibliography

Handbook of Mathematical Functions, Edited by Abramowitz and Stegun, National Bureau of Standards, Applied Mathematics Series, 55 (1964).

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Differentiation of trigonometric functions

Contents

Derivatives of trigonometric functions and their inverses

Proofs of derivative of the sine and cosine functions

Proof of [1]

Differentiating the sine function

Differentiating the cosine function

Proofs of derivatives of inverse trigonometric functions

Differentiating the inverse sine function

Differentiating the inverse cosine function

Differentiating the inverse tangent function

See also

References

Bibliography

Proof of $\lim_{x\to 0}{\frac{\sin(x)}{x}}=1$ ^[1]