Tangent addition formulas

Sine and cosine already add. Tangent and cotangent are quotients of those two. Why is the tangent of a sum not the sum of the tangents, and what denominator repairs the mismatch?

Two angles, one tangent

For all complex $x$ and $y$, away from the poles of the functions that appear,

$$\tan (x\pm y)=\frac{\tan x\pm \tan y}{1\mp \tan x\tan y},$$

$$\cot (x\pm y)=\frac{\cot x\cot y\mp 1}{\cot y\pm \cot x}.$$

The denominator is the debt. Sine and cosine each carry an addition formula, and the quotient of those two expansions is not a sum of quotients.

This picture keeps both angles real and acute, and keeps their sum inside one branch of the tangent. Drag either angle. Two arcs lie on the unit circle. The ray at their sum meets the tangent line, and the addition formula sits on that same cut.

A right-angle shift copies tangent into cotangent, and a half-turn copies each function onto itself up to a sign:

$$\tan \left(\frac{\pi }{2}\pm x\right)=\mp \cot x,$$

$$\cot \left(\frac{\pi }{2}\pm x\right)=\mp \tan x,$$

$$\tan (\pi \pm x)=\pm \tan x,$$

$$\cot (\pi \pm x)=\pm \cot x,$$

$$\tan \left(\frac{3\pi }{2}\pm x\right)=\pm \cot x,$$

$$\cot \left(\frac{3\pi }{2}\pm x\right)=\mp \tan x.$$

Freeze equal angles

Set $y=x$. Away from the poles,

$$\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}=\frac{2}{\cot x-\tan x},$$

$$\cot 2x=\frac{{\cot }^{2}x-1}{2\cot x}=\frac{\cot x-\tan x}{2}.$$

This picture keeps the angle real and small enough that twice the angle stays inside one branch. Drag it. The second arc copies the first. The ray at twice the angle meets the tangent line, and the double-angle formula sits on that cut.

The same addition, applied again, gives three and four times the angle:

$$\tan 3x=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x},$$

$$\cot 3x=\frac{{\cot }^{3}x-3\cot x}{3{\cot }^{2}x-1},$$

$$\tan 4x=\frac{4\tan x-4{\tan }^{3}x}{1-6{\tan }^{2}x+{\tan }^{4}x},$$

$$\cot 4x=\frac{{\cot }^{4}x-6{\cot }^{2}x+1}{4{\cot }^{3}x-4\cot x}.$$

Half the angle

The double-angle relation solves for the half-angle, and the same values rewrite with sine and cosine. Away from the poles,

$$\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x},$$

$$\cot \frac{x}{2}=\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}.$$

This picture keeps $x$ real and below a right angle, so the chord from the leftmost point of the circle still crosses the axis. Drag it. The crossing is $\sin x/(1+\cos x)$. The ray at half the angle meets the tangent line at the same height.

A sum factors

A sum of tangents is a sine over a product of cosines:

$$\tan x\pm \tan y=\frac{\sin (x\pm y)}{\cos x\cos y},$$

$$\cot x\pm \cot y=\pm \frac{\sin (x\pm y)}{\sin x\sin y},$$

$$\tan x+\cot y=\frac{\cos (x-y)}{\cos x\sin y},$$

$$\cot x-\tan y=\frac{\cos (x+y)}{\sin x\cos y}.$$

A product of tangents rewrites against a sum of cotangents:

$$\tan x\tan y=\frac{\tan x+\tan y}{\cot x+\cot y},$$

$$\tan x\tan y=-\frac{\tan x-\tan y}{\cot x-\cot y},$$

$$\cot x\cot y=\frac{\cot x+\cot y}{\tan x+\tan y},$$

$$\cot x\cot y=-\frac{\cot x-\cot y}{\tan x-\tan y},$$

$$\tan x\cot y=\frac{\tan x+\cot y}{\cot x+\tan y},$$

$$\tan x\cot y=-\frac{\tan x-\cot y}{\cot x-\tan y}.$$

Squares convert by the same quotient: ${\tan }^{2}x={\sin }^{2}x/{\cos }^{2}x$, and ${\sin }^{2}x={\tan }^{2}x/(1+{\tan }^{2}x)$. Those two are a lookup once the addition line is in place.

The tangent of a sum carries a denominator built from the product of the tangents. Freezing the two angles gives the double angle, and halving rewrites through sine and cosine. The next pair built from the exponential is hyperbolic sine and cosine.