Mathematics · Ch 3 — Trigonometric Functions

Sum and Difference Formulas for Tangent and Cotangent

The tangent and cotangent sum/difference formulas are not proved from the

unit circle again; they are deduced algebraically from the sine and cosine formulas already

established in Section 6, by dividing through by $\cos x\cos y$ or by $\sin x\sin y$ .

Theorem. For all $x,y$ at which every term is defined,

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

Proof (for $\tan(x+y)$ ). By definition and the Section 6 formulas,

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

Divide numerator and denominator by $\cos x\cos y$ (assumed nonzero):

\tan(x+y) = \frac{\dfrac{\sin x\cos y}{\cos x\cos y}+\dfrac{\cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y}{\cos x\cos y}-\dfrac{\sin x\sin y}{\cos x\cos y}} = \frac{\tan x+\tan y}{1-\tan x\tan y}.

Replacing $y$ by $-y$ (and using $\tan(-y)=-\tan y$ , since $\tan$ is an odd function) gives the

difference case,

\tan(x-y) = \frac{\tan x-\tan y}{1+\tan x\tan y}. \qquad\blacksquare

Theorem. For all $x,y$ at which every term is defined,

\cot(x\pm y) = \frac{\cot x\cot y\mp1}{\cot y\pm\cot x}.

Proof (for $\cot(x+y)$ ). Similarly,

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

This time divide numerator and denominator by $\sin x\sin y$ (assumed nonzero):

\cot(x+y) = \frac{\dfrac{\cos x\cos y}{\sin x\sin y}-1}{\dfrac{\cos y}{\sin y}+\dfrac{\cos x}{\sin x}} = \frac{\cot x\cot y-1}{\cot y+\cot x}.

Replacing $y$ by $-y$ (using $\cot(-y)=-\cot y$ ) gives the difference case,

\cot(x-y) = \frac{\cot x\cot y+1}{\cot y-\cot x}. \qquad\blacksquare

Worked check. With $x=\pi/3,\ y=\pi/4$ : $\tan x=\sqrt3,\ \tan y=1$ , so

$\tan(x+y) = \frac{\sqrt3+1}{1-\sqrt3\cdot1} = \frac{\sqrt3+1}{1-\sqrt3}.$ …