Mathematics · Ch 3 — Trigonometric Functions

Multiple Angle Identities: 2x and 3x

Double-angle identities. Setting $y=x$ in the Section 6 sum formulas

gives, immediately,

\sin 2x = \sin(x+x) = \sin x\cos x+\cos x\sin x = 2\sin x\cos x,

\cos 2x = \cos(x+x) = \cos x\cos x-\sin x\sin x = \cos^2x-\sin^2x.

Using the fundamental identity $\sin^2x+\cos^2x=1$ to eliminate one squared term at a time from

$\cos2x=\cos^2x-\sin^2x$ gives two further, equally useful, forms:

\cos2x = \cos^2x-(1-\cos^2x) = 2\cos^2x-1, \qquad \cos2x = (1-\sin^2x)-\sin^2x = 1-2\sin^2x.

And setting $y=x$ in the Section 7 tangent formula,

\tan2x = \frac{\tan x+\tan x}{1-\tan x\tan x} = \frac{2\tan x}{1-\tan^2x}.

Triple-angle identities. Writing $3x=2x+x$ and applying the sum formulas again, now using

the double-angle results just proved:

\sin3x = \sin(2x+x) = \sin2x\cos x+\cos2x\sin x = (2\sin x\cos x)\cos x + (1-2\sin^2x)\sin x.

Expanding and using $\cos^2x=1-\sin^2x$ on the first term,

\sin3x = 2\sin x(1-\sin^2x) + \sin x - 2\sin^3x = 2\sin x - 2\sin^3x + \sin x - 2\sin^3x = 3\sin x - 4\sin^3x.

Similarly,

\cos3x = \cos(2x+x) = \cos2x\cos x-\sin2x\sin x = (2\cos^2x-1)\cos x - (2\sin x\cos x)\sin x.

Expanding and using $\sin^2x=1-\cos^2x$ on the second term,

\cos3x = 2\cos^3x - \cos x - 2\cos x(1-\cos^2x) = 2\cos^3x-\cos x-2\cos x+2\cos^3x = 4\cos^3x-3\cos x.

Proof of $\tan3x$ . Writing $3x=2x+x$ in the tangent sum formula and substituting

$\tan2x=\dfrac{2\tan x}{1-\tan^2x}$ ,

\tan3x = \frac{\tan2x+\tan x}{1-\tan2x\tan x} = \frac{\dfrac{2\tan x}{1-\tan^2x}+\tan x}{1-\dfrac{2\tan x}{1-\tan^2x}\cdot\tan x}.

Multiplying numerator and denominator through by $(1-\tan^2x)$ :

\text{Numerator} = 2\tan x+\tan x(1-\tan^2x) = 3\tan x-\tan^3x, \qquad \text{Denominator} = (1-\tan^2x)-2\tan^2x = 1-3\tan^2x.

\tan3x = \frac{3\tan x-\tan^3x}{1-3\tan^2x}. \qquad\blacksquare

Summary. …