Differentiation of trigonometric functions

File:limit circle FbN.jpeg

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 < {{sfrac|1|2}} π in the first quadrant.

In the diagram, let R₁ be the triangle OAB, R₂ the circular sector OAB, and R₃ the triangle OAC.

The area of triangle OAB is:

:$\backslash mathrm\{Area\}(R\_1$

)

=\tfrac{1}{2} \ |OA| \ |OB| \sin\theta = \tfrac{1}{2}\sin\theta \, .

The area of the circular sector OAB is:

:$\backslash mathrm\{Area\}(R\_2)$

=\tfrac{1}{2}\theta \, .

The area of the triangle OAC is given by:

:$\backslash mathrm\{Area\}(R\_3$

)

=\tfrac{1}{2} \ |OA| \ |AC| = \tfrac{1}{2} \tan\theta \, .

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

:

\tfrac{1}{2}\sin\theta < \tfrac{1}{2}\theta < \tfrac{1}{2}\tan\theta \, .

Moreover, since {{nowrap|1=sin θ > 0}} in the first quadrant, we may divide through by {{sfrac|1|2}} {{nowrap|sin θ}}, giving:

:

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

File:Squeeze FbN.png

We conclude that for 0 < θ < {{sfrac|1|2}} 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 "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:

$\backslash lim\_\{\backslash theta\; \backslash to\; 0^+\}\; \backslash frac\{\backslash sin\backslash theta\}\{\backslash theta\}\; =\; 1\; \backslash ,\; .$

:$\backslash lim\_\{\backslash theta\; \backslash to\; 0^-\}\backslash !\; \backslash frac\{\backslash sin\backslash theta\}\{\backslash theta\}$

\ =\

\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 \, .

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.

:$\backslash lim\_\{\backslash theta\; \backslash to\; 0\}\backslash ,\; \backslash frac\{\backslash cos\backslash theta\; -\; 1\}\{\backslash theta\}$

\ =\

\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)} .

Using {{nowrap|1=cos²θ – 1 = –sin²θ,}}

the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:

:$\backslash lim\_\{\backslash theta\; \backslash to\; 0\}\backslash ,\backslash frac\{\backslash cos\backslash theta\; -\; 1\}\{\backslash theta\}$

\ =\

\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 \, .

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:

:$$

\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 \, .

We calculate the derivative of the sine function from the limit definition:

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\backslash ,\backslash sin\backslash theta\; =\; \backslash lim\_\{\backslash delta\; \backslash to\; 0\}\; \backslash frac\{\backslash sin(\backslash theta\; +\; \backslash delta)\; -\; \backslash sin\; \backslash theta\}\{\backslash delta\}\; .$

Using the angle addition formula {{nowrap|1=sin(α+β) = sin α cos β + sin β cos α}}, we have:

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\backslash ,\backslash sin\backslash theta$

=

\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) .

Using the limits for the sine and cosine functions:

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\backslash ,\backslash sin\backslash theta$

=

(1)\cos\theta + (0)\sin\theta

=

\cos\theta \, .

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

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\backslash ,\backslash cos\backslash theta$

=

\lim_{\delta \to 0} \frac{\cos(\theta+\delta)-\cos\theta}{\delta} .

Using the angle addition formula {{nowrap|1=cos(α+β) = cos α cos β – sin α sin β}}, we have:

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\backslash ,\backslash cos\backslash theta$

=

\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) .

Using the limits for the sine and cosine functions:

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\backslash ,\backslash cos\backslash theta$

= (0) \cos\theta - (1) \sin\theta = -\sin\theta \, .

To compute the derivative of the cosine function from the chain rule, first observe the following three facts:

:$\backslash cos\backslash theta\; =\; \backslash sin\backslash left(\backslash tfrac\{\backslash pi\}\{2\}-\backslash theta\backslash right)$

:$\backslash sin\backslash theta\; =\; \backslash cos\backslash left(\backslash tfrac\{\backslash pi\}\{2\}-\backslash theta\backslash right)$

:$\backslash tfrac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\; \backslash sin\backslash theta\; =\; \backslash cos\backslash theta$

The first and the second are trigonometric identities, and the third is proven above. Using these three facts, we can write the following,

:$\backslash tfrac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\; \backslash cos\backslash theta\; =\; \backslash tfrac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\; \backslash sin\backslash left(\backslash tfrac\{\backslash pi\}\{2\}-\backslash theta\backslash right)$

We can differentiate this using the chain rule. Letting $f(x)\; =\; \backslash sin\; x,\backslash \; \backslash \; g(\backslash theta)\; =\backslash tfrac\{\backslash pi\}\{2\}-\backslash theta$, we have:

:$\backslash tfrac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\; f\backslash !\backslash left(g\backslash !\backslash left(\backslash theta\backslash right)\backslash right)\; =\; f^\backslash prime\backslash !\backslash left(g\backslash !\backslash left(\backslash theta\backslash right)\backslash right)\; \backslash cdot\; g^\backslash prime\backslash !\backslash left(\backslash theta\backslash right)\; =\; \backslash cos\backslash left(\backslash tfrac\{\backslash pi\}\{2\}-\backslash theta\backslash right)\; \backslash cdot\; (0-1)\; =\; -\backslash sin\backslash theta$.

Therefore, we have proven that

:$\backslash tfrac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\; \backslash cos\backslash theta\; =\; -\backslash sin\backslash theta$.

To calculate the derivative of the tangent function tan θ, we use first principles. By definition:

:$$

\frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\tan\theta

= \lim_{\delta \to 0} \left( \frac{\tan(\theta+\delta)-\tan\theta}{\delta} \right) .

Using the well-known angle formula {{nowrap|1=tan(α+β) = (tan α + tan β) / (1 - tan α tan β)}}, we have:

:$$

\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] .

Using the fact that the limit of a product is the product of the limits:

:$$

\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) .

Using the limit for the tangent function, and the fact that tan δ tends to 0 as δ tends to 0:

:$$

\frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\tan\theta

= 1 \times \frac{1 + \tan^2\theta}{1 - 0} = 1 + \tan^2\theta .

We see immediately that:

:$$

\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 \, .

One can also compute the derivative of the tangent function using the quotient rule.

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\; \backslash tan\backslash theta$

= \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}

The numerator can be simplified to 1 by the Pythagorean identity, giving us,

:$\backslash frac\{1\}\{\backslash cos^2\; \backslash theta\}\; =\; \backslash sec^2\; \backslash theta$

Therefore,

:$\backslash frac\{\backslash operatorname\{d\}\}\{\backslash operatorname\{d\}\backslash !\backslash theta\}\; \backslash tan\backslash theta\; =\; \backslash sec^2\; \backslash theta$

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 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.

We let

:$y=\backslash arcsin\; x\backslash ,\backslash !$

Where

:$-\backslash frac\{\backslash pi\}\{2\}\backslash le\; y\; \backslash le\; \backslash frac\{\backslash pi\}\{2\}$

Then

:$\backslash sin\; y=x\backslash ,\backslash !$

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

:$\{d\; \backslash over\; dx\}\backslash sin\; y=\{d\; \backslash over\; dx\}x$

:$\backslash cos\; y\; \backslash cdot\; \{dy\; \backslash over\; dx\}\; =\; 1\backslash ,\backslash !$

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

:$\backslash sqrt\{1-\backslash sin^2\; y\}\; \backslash cdot\; \{dy\; \backslash over\; dx\}\; =1$

Substituting $x=\backslash sin\; y$ in from above,

:$\backslash sqrt\{1-x^2\}\; \backslash cdot\; \{dy\; \backslash over\; dx\}\; =1$

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

We let

:$y=\backslash arccos\; x\backslash ,\backslash !$

Where

:$0\; \backslash le\; y\; \backslash le\; \backslash pi$

Then

:$\backslash cos\; y=x\backslash ,\backslash !$

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

:$\{d\; \backslash over\; dx\}\backslash cos\; y=\{d\; \backslash over\; dx\}x$

:$-\backslash sin\; y\; \backslash cdot\; \{dy\; \backslash over\; dx\}\; =1$

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

:$-\backslash sqrt\{1-\backslash cos^2\; y\}\; \backslash cdot\; \{dy\; \backslash over\; dx\}\; =1$

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

:$-\backslash sqrt\{1-x^2\}\; \backslash cdot\; \{dy\; \backslash over\; dx\}\; =1$

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

Alternatively, once the derivative of $\backslash arcsin\; x$ is established, the derivative of $\backslash arccos\; x$ follows immediately by differentiating the identity $\backslash arcsin\; x+\backslash arccos\; x=\backslash pi/2$ so that $(\backslash arccos\; x)\text{'}=-(\backslash arcsin\; x)\text{'}$.

We let

:$y=\backslash arctan\; x\backslash ,\backslash !$

Where

:

Then

:$\backslash tan\; y=x\backslash ,\backslash !$

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

:$\{d\; \backslash over\; dx\}\backslash tan\; y=\{d\; \backslash over\; dx\}x$

Left side:

:$$

{d \over dx}\tan y

= \sec^2 y \cdot {dy \over dx}

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

using the Pythagorean identity

Right side:

:$\{d\; \backslash over\; dx\}x\; =\; 1$

Therefore,

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

Substituting $x=\backslash tan\; y\backslash ,\backslash !$ in from above, we get

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

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

We let

:$y=\backslash arccot\; x$

where

:$\backslash cot\; y=x$

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

:$\backslash frac\{d\}\{dx\}\backslash cot\; y=\backslash frac\{d\}\{dx\}x$

Left side:

:$$

{d \over dx}\cot y

= -\csc^2 y \cdot {dy \over dx}

= -(1 + \cot^2 y) {dy \over dx}

using the Pythagorean identity

Right side:

:$\{d\; \backslash over\; dx\}x\; =\; 1$

Therefore,

:$-(1+\backslash cot^2y)\backslash frac\{dy\}\{dx\}=1$

Substituting $x=\backslash cot\; y$,

:$-(1+x^2)\backslash frac\{dy\}\{dx\}=1$

:$\backslash frac\{dy\}\{dx\}=-\backslash frac\{1\}\{1+x^2\}$

Alternatively, as the derivative of $\backslash arctan\; x$ is derived as shown above, then using the identity $\backslash arctan\; x+\backslash arccot\; x=\backslash dfrac\{\backslash pi\}\{2\}$ follows immediately that$$\backslash begin\{align\}$$

\dfrac{d}{dx}\arccot x

&=\dfrac{d}{dx}\left(\dfrac{\pi}{2}-\arctan x\right)\\

&=-\dfrac{1}{1+x^2}

\end{align}

Let

:$y\; =\; \backslash arcsec\; x\backslash \; \backslash mid\; |x|\; \backslash geq\; 1$

Then

:$x\; =\; \backslash sec\; y\; \backslash mid\; \backslash \; y\; \backslash in\; \backslash left\; [0,\backslash frac\{\backslash pi\}\{2\}\; \backslash right\; )\backslash cup\; \backslash left\; (\backslash frac\{\backslash pi\}\{2\},\backslash pi\; \backslash right]$

:$\backslash frac\{dx\}\{dy\}\; =\; \backslash sec\; y\; \backslash tan\; y\; =\; |x|\backslash sqrt\{x^2-1\}$

(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 $\backslash sqrt\{x^2-1\}$ 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.)

:$\backslash frac\{dy\}\{dx\}\; =\; \backslash frac\{1\}\{|x|\backslash sqrt\{x^2-1\}\}$

Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using the chain rule.

Let

:$y\; =\; \backslash arcsec\; x\; =\; \backslash arccos\; \backslash left(\backslash frac\{1\}\{x\}\backslash right)$

Where

:$|x|\; \backslash geq\; 1$ and $y\; \backslash in\; \backslash left[0,\; \backslash frac\{\backslash pi\}\{2\}\backslash right)\; \backslash cup\; \backslash left(\backslash frac\{\backslash pi\}\{2\},\; \backslash pi\backslash right]$

Then, applying the chain rule to $\backslash arccos\; \backslash left(\backslash frac\{1\}\{x\}\backslash right)$:

:$\backslash frac\{dy\}\{dx\}\; =\; -\backslash frac\{1\}\{\backslash sqrt\{1-(\backslash frac\{1\}\{x\})^2\}\}\; \backslash cdot\; \backslash left(-\backslash frac\{1\}\{x^2\}\backslash right)$

= \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}}

Let

:$y\; =\; \backslash arccsc\; x\backslash \; \backslash mid\; |x|\; \backslash geq\; 1$

Then

:$x\; =\; \backslash csc\; y\backslash \; \backslash mid\; \backslash \; y\; \backslash in\; \backslash left\; [-\backslash frac\{\backslash pi\}\{2\},0\; \backslash right\; )\backslash cup\; \backslash left\; (0,\backslash frac\{\backslash pi\}\{2\}\; \backslash right]$

:$\backslash frac\{dx\}\{dy\}\; =\; -\backslash csc\; y\; \backslash cot\; y\; =\; -|x|\backslash sqrt\{x^2-1\}$

(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 $\backslash sqrt\{x^2-1\}$ 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.)

:$\backslash frac\{dy\}\{dx\}\; =\; \backslash frac\{-1\}\{|x|\backslash sqrt\{x^2-1\}\}$

Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using the chain rule.

Let

:$y\; =\; \backslash arccsc\; x\; =\; \backslash arcsin\; \backslash left(\backslash frac\{1\}\{x\}\backslash right)$

Where

:$|x|\; \backslash geq\; 1$ and $y\; \backslash in\; \backslash left[-\backslash frac\{\backslash pi\}\{2\},\; 0\backslash right)\; \backslash cup\; \backslash left(0,\; \backslash frac\{\backslash pi\}\{2\}\backslash right]$

Then, applying the chain rule to $\backslash arcsin\; \backslash left(\backslash frac\{1\}\{x\}\backslash right)$:

:$\backslash frac\{dy\}\{dx\}\; =\backslash frac\{1\}\{\backslash sqrt\{1-(\backslash frac\{1\}\{x\})^2\}\}\; \backslash cdot\; \backslash left(-\backslash frac\{1\}\{x^2\}\backslash right)$

= -\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}}

• {{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}}

{{Reflist}}

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

{{DEFAULTSORT:Differentiation Of Trigonometric Functions}}

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Category:Differential calculus

Category:Mathematical identities