Why the Inverse Tangents of 1, 2 and 3 Add Up to Exactly Pi

Just as sine and cosine obey identities that simplify expressions, their inverses have identities of their own. These let a tangled sum of inverse tangents collapse to a single angle, or turn an equation in ${\sin }^{-1}$ and ${\cos }^{-1}$ into ordinary algebra.

This part covers co-function identities, the inverse tangent addition formulae, double and triple angle forms, and solving equations.

**For every x in $[-1,1]$, ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$; similarly ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$ for all real x, and ${\sec }^{-1}x+{\text{cosec}}^{-1}x=\frac{\pi }{2}$ for $|x|\ge 1$.

Proof.** Let $\theta ={\sin }^{-1}x$, so $x=\sin \theta =\cos \left(\frac{\pi }{2}-\theta \right)$. Since $\theta$ lies in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, the angle $\frac{\pi }{2}-\theta$ lies in $[0,\pi ]$, the principal branch of ${\cos }^{-1}$. So ${\cos }^{-1}x=\frac{\pi }{2}-\theta$, and adding gives $\frac{\pi }{2}$.

Worked example. If ${\sin }^{-1}x=\frac{\pi }{5}$, then ${\cos }^{-1}x=\frac{\pi }{2}-\frac{\pi }{5}=\frac{3\pi }{10}$.

Numerical check. ${\sin }^{-1}0.6\approx 0.6435$ and ${\cos }^{-1}0.6\approx 0.9273$, whose sum is $1.5708\approx \frac{\pi }{2}$.

An everyday example. A ladder leaning against a wall makes one angle with the ground and another with the wall; one is the inverse sine and the other the inverse cosine of the same ratio, and together they form a right angle.

The substance. The proof works only because the branches fit together — $\frac{\pi }{2}-\theta$ lands exactly in $[0,\pi ]$, so the identity holds for negative x as well.

**For $xy<1$, ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$, and for $xy>-1$, ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\frac{x-y}{1+xy}$; when $x>0$, $y>0$ and $xy>1$, the sum becomes $\pi +{\tan }^{-1}\frac{x+y}{1-xy}$.

Proof.** Let $A={\tan }^{-1}x$ and $B={\tan }^{-1}y$. Then

$$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{x+y}{1-xy}$$

When $xy<1$, $A+B$ stays inside $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, so taking ${\tan }^{-1}$ of both sides is valid.

Worked example. ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}={\tan }^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}={\tan }^{-1}1=\frac{\pi }{4}$.

**Worked example 2, where $xy>1$.** ${\tan }^{-1}2+{\tan }^{-1}3=\pi +{\tan }^{-1}\frac{5}{1-6}=\pi +{\tan }^{-1}(-1)=\frac{3\pi }{4}$. Adding ${\tan }^{-1}1=\frac{\pi }{4}$ gives ${\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\pi$.

An everyday example. A 10 m flagpole on top of a 20 m building, seen from 30 m away, makes an angle at your eye of ${\tan }^{-1}1-{\tan }^{-1}\frac{2}{3}={\tan }^{-1}\frac{1-\frac{2}{3}}{1+\frac{2}{3}}={\tan }^{-1}\frac{1}{5}\approx 11.3°$.

The substance. **Ignoring the condition $xy<1$ gives impossible answers** — the plain formula for ${\tan }^{-1}2+{\tan }^{-1}3$ would give $-\frac{\pi }{4}$, although both angles are positive.

**The double angle forms include $2{\tan }^{-1}x={\sin }^{-1}\frac{2x}{1+x^2}={\cos }^{-1}\frac{1-x^2}{1+x^2}={\tan }^{-1}\frac{2x}{1-x^2}$, and the triple angle forms include $3{\sin }^{-1}x={\sin }^{-1}(3x-4x^3)$ and $3{\cos }^{-1}x={\cos }^{-1}(4x^3-3x)$, each valid on a stated interval.

Results with their conditions:**

- $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$ for $|x|<1$
- $2{\tan }^{-1}x={\sin }^{-1}\frac{2x}{1+x^2}$ for $|x|\le 1$
- $2{\tan }^{-1}x={\cos }^{-1}\frac{1-x^2}{1+x^2}$ for $x\ge 0$
- $3{\sin }^{-1}x={\sin }^{-1}(3x-4x^3)$ for $|x|\le \frac{1}{2}$
- $3{\cos }^{-1}x={\cos }^{-1}(4x^3-3x)$ for $\frac{1}{2}\le x\le 1$

Where they come from. Put $x=\tan \theta$: then $\frac{2x}{1+x^2}=\sin 2\theta$, so ${\sin }^{-1}\frac{2x}{1+x^2}=2\theta =2{\tan }^{-1}x$.

Worked example. For $x=\frac{1}{2}$, ${\sin }^{-1}\frac{2x}{1+x^2}={\sin }^{-1}0.8\approx 0.9273$, and $2{\tan }^{-1}\frac{1}{2}\approx 0.9273$ as well.

Worked example 2. $3{\sin }^{-1}\frac{1}{2}={\sin }^{-1}\left(\frac{3}{2}-\frac{4}{8}\right)={\sin }^{-1}1=\frac{\pi }{2}$, matching $3\times \frac{\pi }{6}$.

An everyday example. **Doubling the tilt of a rooftop solar panel set at ${\tan }^{-1}\frac{1}{3}$** gives ${\tan }^{-1}\frac{2/3}{1-1/9}={\tan }^{-1}\frac{3}{4}$, about $36.9°$.

The substance. Each formula holds only on its interval — outside it, extra multiples of $\pi$ appear.

Equations are solved by combining the inverse terms with the identities, taking the trigonometric function of both sides to reach an algebraic equation, and rejecting any root that falls outside the domains or branches involved.

Worked example. Solve ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$.

$$\frac{2x+3x}{1-6x^2}=1\Rightarrow 6x^2+5x-1=0\Rightarrow (6x-1)(x+1)=0$$

So $x=\frac{1}{6}$ or $x=-1$. For $x=-1$ both angles are negative and cannot add to $\frac{\pi }{4}$, so the only solution is $x=\frac{1}{6}$. Check: ${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{2}=\frac{\pi }{4}$.

Worked example 2. Solve ${\sin }^{-1}x=2{\cos }^{-1}x$. Using ${\sin }^{-1}x=\frac{\pi }{2}-{\cos }^{-1}x$ gives $\frac{\pi }{2}=3{\cos }^{-1}x$, so ${\cos }^{-1}x=\frac{\pi }{6}$ and $x=\frac{\sqrt{3}}{2}$.

Worked example 3. For $-\frac{\pi }{2}<x<\frac{\pi }{2}$, ${\tan }^{-1}\frac{\cos x}{1+\sin x}=\frac{\pi }{4}-\frac{x}{2}$; at $x=0$ both sides equal $\frac{\pi }{4}$.

An everyday example. Finding how far to stand so that a statue and its pedestal look equally tall leads to exactly this kind of inverse tangent equation.

The substance. Always test roots in the original equation — taking tan of both sides can introduce roots that do not satisfy it.