Calculus – Differentiation – Remarkable identities

We have seen that \(\displaystyle\frac{d}{dx}(\arcsin(x))=\displaystyle\frac{1}{\sqrt{1-x^2}}\) and \(\displaystyle\frac{d}{dx}(\arccos(x))=-\displaystyle\frac{1}{\sqrt{1-x^2}}\). Then we have:

\[\frac{d}{dx}\left(\arcsin(x)+\arccos(x)\right)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0,\quad -1 < x < 1.\]

This implies that \(\arcsin(x)+\arccos(x)\) is a constant function for \(-1 < x < 1\). Since \(\arcsin(0)+\arccos(0)=0+\frac{1}{2}\pi=\frac{1}{2}\pi\) it follows that \(\arcsin(x)+\arccos(x)=\frac{1}{2}\pi\) for \(-1 < x < 1\). However, since we also have

\[\arcsin(-1)+\arccos(-1)=-\tfrac{1}{2}\pi+\pi=\tfrac{1}{2}\pi\quad\text{and}\quad\arcsin(1)+\arccos(1)=\tfrac{1}{2}\pi+0=\tfrac{1}{2}\pi,\]

we conclude that

\[\arcsin(x)+\arccos(x)=\tfrac{1}{2}\pi,\quad -1\leq x\leq 1.\]

Another example is based on \(y=\arctan\left(\frac{1}{x}\right)\) for \(x\neq0\). Using the chain rule we obtain that

\[\frac{dy}{dx}=\frac{1}{1+\left(\frac{1}{x}\right)^2}\cdot\left(-\frac{1}{x^2}\right)=-\frac{1}{1+x^2},\quad x\neq0.\] This implies that the function \(\arctan(x)+\arctan\left(\frac{1}{x}\right)\) is constant for \(x>0\) and for \(x<0\).

Since \(\arctan(1)=\frac{1}{4}\pi\) and \(\arctan(-1)=-\frac{1}{4}\pi\), we conclude that

\[\arctan(x)+\arctan\left(\tfrac{1}{x}\right)=\tfrac{1}{2}\pi,\quad x>0\]

and

\[\arctan(x)+\arctan\left(\tfrac{1}{x}\right)=-\tfrac{1}{2}\pi,\quad x<0.\]

Stewart §3.5, Exercise 54
Find the derivative of \(y=\arctan\left(x-\sqrt{1+x^2}\right)\).

Solution:
Using the chain rule we obtain that

\begin{align*} \frac{dy}{dx}&=\frac{1}{1+\left(x-\sqrt{1+x^2}\right)^2}\cdot\left(1-\frac{2x}{2\sqrt{1+x^2}}\right)=\frac{1-\frac{x}{\sqrt{1+x^2}}}{1+x^2-2x\sqrt{1+x^2}+1+x^2}\\[2.5mm] &=\frac{\sqrt{1+x^2}-x}{2\sqrt{1+x^2}(1+x^2-x\sqrt{1+x^2})}=\frac{\sqrt{1+x^2}-x}{2(1+x^2)(\sqrt{1+x^2}-x)}=\frac{1}{2(1+x^2)}. \end{align*}

This result is remarkable, because this equals the derivative of \(\arctan(x)\) except for a factor \(2\). Note that

\[f(x)=\arctan(x)-2\arctan\left(x-\sqrt{1+x^2}\right)\quad\Longrightarrow\quad f'(x)=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0.\]

This implies that \(f(x)\) is constant. Now we have \(f(0)=\arctan(0)-2\arctan(-1)=0+\tfrac{1}{2}\pi=\tfrac{1}{2}\pi\). Hence we have:

\[\arctan(x)-2\arctan\left(x-\sqrt{1+x^2}\right)=\tfrac{1}{2}\pi,\quad x\in\mathbb{R}.\]

Stewart §3.5, Exercise 60
Find the derivative of \(y=\arctan\left(\sqrt{\displaystyle\frac{1-x}{1+x}}\right)\) for \(x\geq0\).

Solution:
Using the chain rule we obtain that

\[\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\displaystyle\frac{1-x}{1+x}}\right)^2}\cdot\frac{1}{\sqrt{\displaystyle\frac{1-x}{1+x}}}\cdot\frac{-(1+x)-(1-x)}{(1+x)^2} =-\frac{1}{2\sqrt{1-x^2}},\quad x\geq0,\quad x\neq 1.\]

This implies that the function \(g(x)=\arcsin(x)+2\arctan\left(\sqrt{\displaystyle\frac{1-x}{1+x}}\right)\) is constant on its domain \([0,1]\).

Since for instance we have

\[g(0)=\arcsin(0)+2\arctan(1)=\tfrac{1}{2}\pi\quad\text{or}\quad g(1)=\arcsin(1)+2\arctan(0)=\tfrac{1}{2}\pi,\]

we conclude that

\[\arcsin(x)+2\arctan\left(\sqrt{\frac{1-x}{1+x}}\right)=\tfrac{1}{2}\pi,\quad 0\leq x\leq 1.\]

Stewart §4.2, Exercise 34
Prove the identity \(2\arcsin(x)=\arccos\left(1-2x^2\right)\) for \(0\leq x\leq1\).

Solution:
Consider the function \(h(x)=2\arcsin(x)-\arccos\left(1-2x^2\right)\) for \(0\leq x\leq 1\). Then we have:

\[h'(x)=\frac{2}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(1-2x^2)^2}}\cdot(-4x)=\frac{2}{\sqrt{1-x^2}}-\frac{4x}{\sqrt{1-1+4x^2-4x^4}} =\frac{2}{\sqrt{1-x^2}}-\frac{4x}{2x\sqrt{1-x^2}}=0.\]

Hence: \(h(x)\) is constant for \(0\leq x\leq 1\). Now we have

\[h(0)=2\arcsin(0)-\arccos(1)=0-0=0\quad\text{and}\quad h(1)=2\arcsin(1)-\arccos(-1)=\pi-\pi=0.\]

Hence: \(2\arcsin(x)=\arccos\left(1-2x^2\right)\) for \(0\leq x\leq1\).

Stewart §4.2, Exercise 35
Prove the identity \(\arcsin\left(\displaystyle\frac{x-1}{x+1}\right)=2\arctan\left(\sqrt{x}\right)-\frac{1}{2}\pi\) for \(x\geq0\).

Solution:
Suppose that \(k(x)=\arcsin\left(\displaystyle\frac{x-1}{x+1}\right)-2\arctan\left(\sqrt{x}\right)\) for \(x\geq0\). Then we have:

\begin{align*} k'(x)&=\frac{1}{\sqrt{1-\left(\displaystyle\frac{x-1}{x+1}\right)^2}}\cdot\frac{x+1-(x-1)}{(x+1)^2}-\frac{2}{1+x}\cdot\frac{1}{2\sqrt{x}}\\[2.5mm] &=\frac{2}{(x+1)\sqrt{(x+1)^2-(x-1)^2}}-\frac{1}{(x+1)\sqrt{x}}=\frac{1}{(x+1)\sqrt{x}}-\frac{1}{(x+1)\sqrt{x}}=0. \end{align*}

Hence: \(k(x)\) is constant for \(x\geq0\). Now we have \(k(0)=\arcsin(-1)-2\arctan(0)=-\frac{1}{2}\pi-0=-\frac{1}{2}\pi\). This implies that \(\arcsin\left(\displaystyle\frac{x-1}{x+1}\right)=2\arctan\left(\sqrt{x}\right)-\frac{1}{2}\pi\) for \(x\geq0\).

If you want to do it yourself:

Exercises:

1. Prove the identity \(\displaystyle\arcsin(x)=\arccos\left(\sqrt{1-x^2}\right)\) for \(0\leq x\leq 1\).


2. Prove the identity \(\displaystyle\arcsin(x)=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)\) for \(-1\leq x\leq 1\).


3. Prove the identity \(\displaystyle\arctan(x)=\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)\) for all \(x\in\mathbb{R}\).


4. Prove the identity \(\displaystyle\arctan(x)=\arccos\left(\frac{1}{\sqrt{1+x^2}}\right)\) for all \(x\in\mathbb{R}\).

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Last modified on March 1, 2021