Trigonometric functions

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In this article, we present the six usual trigonometric functions, their derivatives, and the derivatives of their inverse.

$$\DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\arccsch}{arccsch} \sinh x$$ $$\arccsch x$$

Sine and cosine

$$\sin x$$ $$\cos x$$

Tangent

$$\tan x=\frac{\sin x}\cos{x}$$

Cotangent

$$\cot x=\frac{1}{\tan x}$$

Secant

$$\sec x=\frac{1}{\cos x}$$

Cosecant

$$\csc x=\frac{1}{\sin x}$$

Derivatives

Derivative of the sine

$$\sin^\prime x=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$$

Using $\sin(a+b)=\sin a\cos b+\cos a\sin b$, we get

$$\sin^\prime x=\lim_{h\to0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$$

Then rearranging

$$\sin^\prime x=\sin x\left(\lim_{h\to0}\frac{\cos h-1}{h}\right)+\cos x\left(\lim_{h\to0}\frac{\sin h}{h}\right)$$

Using $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ and $\displaystyle\lim_{x\to0}\frac{\cos x-1}{x}=0$, we get

$$\fbox{$\sin^\prime x=\cos x$}$$

Derivative of the cosine

To compute the derivative of the cosine, we use the identity $\cos^2 x+\sin^2 x=1$

$$\cos^\prime x=\frac{d\sqrt{1-\sin^2 x}}{dx}$$

This is only valid for those $x$ for which $\cos x$ is positive. Applying the chain rule, we get

$$\cos^\prime x=\frac{1}{2\sqrt{1-\sin^2 x}}\left(-2\sin x\right)\cos x$$

Simplifying, we get

$$\fbox{$\cos^\prime x=-\sin x$}$$

If $\cos x$ is negative, then $\cos x=-\sqrt{1-\sin^2 x}$, and

$$\cos^\prime x=-\frac{1}{2\sqrt{1-\sin^2 x}}(-2\sin x)\cos x$$

which yields

$$\cos^\prime x=-\frac{1}{2(-\cos x)}(-2\sin x)\cos x$$

After simplification, we get the same result

$$\fbox{$\cos^\prime x=-\sin x$}$$

Derivative of the tangent

To derive the tangent, we simply use the derivation of quotients

$$\tan^\prime x=\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)=\frac{\sin^\prime x\cos x-\sin x\cos^\prime x}{\cos^2 x}=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}=\sec^2 x$$ $$\fbox{$\tan^\prime x=\sec^2 x$}$$

Derivative of the cotangent

To derive the tangent, we use the derivation of the inverse

$$\cot^\prime x=\frac{d}{dx}\left(\frac{1}{\tan x}\right)=-\tan^\prime x\frac{1}{\tan^2 x}=-\sec^2 x\frac{\cos^2 x}{\sin^2 x}=-\frac{1}{\sin^2 x}=-\csc^2 x$$ $$\fbox{$\cot^\prime x=-\csc^2 x$}$$

Derivative of the secant

To derive the secant, we use the derivation of the inverse

$$\sec^\prime x=\frac{d}{dx}\left(\frac{1}{\cos x}\right)=-\cos^\prime x\frac{1}{\cos^2 x}=\frac{\sin x}{\cos^2 x}=\sec x\tan x$$ $$\fbox{$\sec^\prime x=\sec x\tan x$}$$

Derivative of the cosecant

To derive the cosecant, we use the derivation of the inverse

$$\csc^\prime x=\frac{d}{dx}\left(\frac{1}{\sin x}\right)=-\sin^\prime x\frac{1}{\sin^2 x}=-\frac{\cos x}{\sin^2 x}=-\csc x\cot x$$ $$\fbox{$\csc^\prime x=-\csc x\cot x$}$$

Derivatives of the inverse trigonometric functions

To derive the inverse function, we use the following identity

$$\left(f^{-1}\right)^\prime(x)=\frac{1}{f^\prime(f^{-1}(x))}$$

Derivative of the inverse sine

$$\arcsin^\prime x=\frac{1}{\sin^\prime(\arcsin x))}=\frac{1}{\cos(\arcsin x))}$$

Therefore, when $\cos x$ is positive,

$$\arcsin^\prime (\sin x)=\frac{1}{\cos(\arcsin (\sin x)))}=\frac{1}{\cos x}=\frac{1}{\sqrt{1-\sin^2 x}}$$

Substituting $\sin x$ with $x$, we get

$$\fbox{$\arcsin^\prime x=\frac{1}{\sqrt{1-x^2}}$}$$

$\cos x$ is positive from $-\pi/2$ to $\pi/2$, but on this interval, $\sin x$ goes from $-1$ to $1$, i.e. $\sin x$ spans the whole range of the sine function, which is the whole domain of the inverse sine function.

Therefore, we do not need to consider the case where $\cos x$ is negative because we assume the range of $\arcsin$ to be $[-\pi/2,\pi/2]$.

Derivative of the inverse cosine

$$\arccos^\prime x=\frac{1}{\cos^\prime(\arccos x))}=-\frac{1}{\sin(\arccos x))}$$

Therefore, when $\sin x$ is positive,

$$\arccos^\prime (\cos x)=-\frac{1}{\sin(\arccos (\cos x)))}=-\frac{1}{\sin x}=-\frac{1}{\sqrt{1-\cos^2 x}}$$

Substituting $\cos x$ with $x$, we get

$$\fbox{$\arccos^\prime x=-\frac{1}{\sqrt{1-x^2}}$}$$

$\sin x$ is positive on $[0,\pi]$, and on this interval, $\cos x$ goes from $1$ to $-1$, so we do not need to consider the case where $\sin x$ is negative when we define the range of $\arccos x$ to be $[0,\pi]$.

Derivative of the inverse tangent

$$\arctan^\prime x=\frac{1}{\tan^\prime(\arctan x)}=\frac{1}{\sec^2(\arctan x)}$$

Therefore,

$$\arctan^\prime (\tan x)=\frac{1}{\sec^2(\arctan (\tan x))}=\frac{1}{\sec^2 x}$$

Now, we need to express $\tan x$ as a function of $\sec^2 x$

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

Which yields

$$\sec^2 x=1+\tan^2 x$$

Therefore

$$\arctan^\prime (\tan x)=\frac{1}{\sec^2 x}=\frac{1}{1+\tan^2 x}$$

Substituting $\tan x$ with $x$, we get

$$\fbox{$\arctan^\prime x=\frac{1}{1+x^2}$}$$

Derivative of the inverse cotangent

The computation is quite similar to that of the inverse tangent.

$$\arccot^\prime x=\frac{1}{\cot^\prime(\arccot x)}=-\frac{1}{\csc^2(\arccot x)}$$

Therefore

$$\arccot^\prime (\cot x)=-\frac{1}{\csc^2(\arccot (\cot x))}=-\frac{1}{\csc^2 x}$$

We need to express $\csc^2 x$ as a function of $\cot x$.

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

which yields

$$\csc^2x=1+\cot^2 x$$

Therefore

$$\arccot^\prime (\cot x)=-\frac{1}{\csc^2 x}=-\frac{1}{1+\cot^2 x}$$

and substituting $\cot x$ with $x$, we get

$$\fbox{$\arccot^\prime x=-\frac{1}{1+x^2}$}$$

Derivation of the inverse secant

$$\arcsec^\prime x=\frac{1}{\sec^\prime(\arcsec x)}=\frac{1}{\sec(\arcsec x)\tan(\arcsec x)}$$

Therefore

$$\arcsec^\prime (\sec x)=\frac{1}{\sec(\arcsec (\sec x))\tan(\arcsec (\sec x))}=\frac{1}{\sec x\tan x}$$

We need to rewrite $\sec x\tan x$ as a function of $\sec x$ only. Assuming $\sin x$ positive,

$$\sec x\tan x=\sec x\frac{\sin x}{\cos x}=\sec^2 x\sin x=\sec^2 x\sqrt{1-\cos^2 x}=\sec^2 x\sqrt{1-\frac{1}{\sec^2 x}}$$

Therefore,

$$\arcsec^\prime (\sec x)=\frac{1}{\sec^2 x\sqrt{1-\frac{1}{\sec^2 x}}}$$

and substituting $\sec x$ with $x$, we get

$$\fbox{$\arcsec^\prime x=\frac{1}{x^2 \sqrt{1-\frac{1}{x^2}}}$}$$

If we take the range of $\sec x$ to be $[0,\pi/2]$, then the case $\sin x$ negative never occurs.

Derivation of the inverse cosecant

$$\arccsc^\prime x=\frac{1}{\csc^\prime(\arccsc x)}=-\frac{1}{\csc(\arccsc x)\cot(\arccsc x)}$$

Therefore

$$\arccsc^\prime (\csc x)=-\frac{1}{\csc(\arccsc (\csc x))\cot(\arccsc (\csc x))}=-\frac{1}{\csc x\cot x}$$

We need to rewrite $\csc x\cot x$ as a function of $\csc x$ only. Assuming $\csc x$ positive,

$$\csc x\cot x=\csc x\frac{\cos x}{\sin x}=\csc^2 x\cos x=\csc^2 x\sqrt{1-\sin^2 x}=\csc^2 x\sqrt{1-\frac{1}{\csc^2 x}}$$

Therefore

$$\arccsc^\prime (\csc x)=-\frac{1}{\csc^2 x\sqrt{1-\frac{1}{\csc^2 x}}}$$

And substituting $\csc x$ with $x$, we get

$$\fbox{$\arccsc^\prime x=-\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}$}$$

Summary

So far, we know

$$\arcsin^\prime x=\frac{1}{\sqrt{1-x^2}}$$ $$\arccos^\prime x=-\frac{1}{\sqrt{1-x^2}}$$ $$\arctan^\prime x=\frac{1}{1+x^2}$$ $$\arccot^\prime x=-\frac{1}{1+x^2}$$ $$\arcsec^\prime x=\frac{1}{x^2 \sqrt{1-\frac{1}{x^2}}}$$ $$\arccsc^\prime x=-\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}$$

Primitive of the inverse trigonometric functions

To compute the primitive of a function, we use the following identity, which we derive using integration by parts

$$\int f^{-1}(x)dx=x f^{-1}(x)-\int x \left(f^{-1}\right)^\prime(x)dx$$

For readability, we omit the integration constant in all the results.

Primitive of the inverse sine

$$\int\arcsin x dx=x\arcsin x-\int x\arcsin^\prime x dx$$ $$\int x\arcsin^\prime x dx=\int\frac{x}{\sqrt{1-x^2}}dx$$

With $u=1-x^2$, we have $du=-2xdx$ and $dx=-\frac{du}{2}$, which yields

$$\int x\arcsin^\prime x dx=\int\frac{x}{\sqrt{1-x^2}}dx=-\frac{1}{2}\int\frac{du}{\sqrt{u}}=-\frac{1}{2}\left(\frac{2}{1}\sqrt{u}\right)=-\sqrt{u}=-\sqrt{1-x^2}$$

Therefore,

$$\fbox{$\int\arcsin x dx=x\arcsin x+\sqrt{1-x^2}$}$$

Primitive of the inverse cosine

The computation is very similar to that of the inverse sine.

$$\int\arccos x dx=x\arccos x-\int x\arccos^\prime x dx$$ $$\int x\arccos^\prime x dx=-\int\frac{x}{\sqrt{1-x^2}}dx=\sqrt{1-x^2}$$

Therefore,

$$\fbox{$\int\arccos x dx=x\arccos x-\sqrt{1-x^2}$}$$

Primitive of the inverse tangent

$$\int\arctan x dx=x\arctan x-\int x\arctan^\prime x dx$$ $$\int x\arctan^\prime x dx=\int \frac{x}{1+x^2} dx$$

With $u=1+x^2$, we have $du=2 x dx$ and $x dx=\frac{du}{2}$, which yields

$$\int \frac{x}{1+x^2} dx=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|=\frac{1}{2}\ln|1+x^2|=\ln\sqrt{1+x^2}$$

Therefore,

$$\fbox{$\int\arctan x dx=x\arctan x-\ln\sqrt{1+x^2}$}$$

Primitive of the inverse cotangent

The computation is very similar to that of the inverse tangent.

$$\int\arccot x dx=x\arccot x-\int x\arccot^\prime x dx$$ $$\int x\arccot^\prime x dx=-\int\frac{x}{1+x^2}dx=-\ln\sqrt{1+x^2}$$

Therefore,

$$\fbox{$\int\arccot x dx=x\arccot x+\ln\sqrt{1+x^2}$}$$

Primitive of the inverse secant

$$\int\arcsec x dx=x\arcsec x-\int x\arcsec^\prime x dx$$ $$\int x\arcsec^\prime x dx=\int\frac{x}{x^2 \sqrt{1-\frac{1}{x^2}}}dx=\int\frac{1}{x \sqrt{1-\frac{1}{x^2}}}dx$$