Multiple angle identities and submultiple angle identities

Multiple angles of $A$ are $2A, 3A, 4A,\ldots$ -- the angle scaled up by a whole number -- while sub-multiple angles are $\dfrac{A}{2},\dfrac{A}{3},\ldots$ -- the angle scaled down. This section turns the compound-angle formulas of the previous section into identities that express $\sin2A$, $\cos3A$, $\tan\dfrac{\theta}{2}$, and so on, purely through the ratios of the single angle $A$ (or $\theta$) itself.

Double-Angle Identities

Putting $\beta=\alpha$ into the sum formulas of the previous section collapses them into formulas for $2\alpha$:

Identity 3.7. From $\sin(A+A)=\sin A\cos A+\cos A\sin A$:

Identity 3.8. From $\cos(A+A)=\cos A\cos A-\sin A\sin A$:

Substituting the Pythagorean identity $\sin^2A=1-\cos^2A$ or $\cos^2A=1-\sin^2A$ gives two more equivalent all-one-ratio forms, all three worth memorising:

Identity 3.9. From $\tan(A+A)=\dfrac{\tan A+\tan A}{1-\tan A\tan A}$:

Identities 3.10 & 3.11 (tan-only forms of sine/cosine). Dividing $\sin2A=2\sin A\cos A$ top and bottom by $\cos^2A+\sin^2A=1$, then by $\cos^2A$, rewrites $\sin2A$ purely in terms of $\tan A$; the same trick on $\cos2A=\cos^2A-\sin^2A$ gives $\cos2A$ in terms of $\tan A$:

These are handy when $A$'s tangent, not its sine or cosine, is what is given.

Worked illustration (projectile range). A projectile launched at speed $u$ and angle $\alpha$ over level ground travels a horizontal range $R=\dfrac{u^2\sin2\alpha}{g}$ before landing -- this single formula is why the range is greatest exactly at $\alpha=45^\circ$, where $\sin2\alpha=\sin90^\circ=1$ is as large as it can get. For a football kicked at $u=80$ ft/s with $g=32$ ft/s$^2$: $R=\dfrac{80^2\sin2\alpha}{32}=200\sin2\alpha$, so the maximum possible distance is $200$ ft, reached by kicking at $45^\circ$. The same identity, written as $\sin A\cos A=\tfrac12\sin2A$, shows this product never leaves $\left[-\tfrac12,\tfrac12\right]$, with the top value $\tfrac12$ achieved exactly at $A=\tfrac{\pi}{4}$.

Power-Reducing (Reduction) Identities

Solving the all-cosine double-angle form for the squared ratio runs the double-angle identity backwards: instead of writing $\cos2A$ in terms of $\cos^2A$, it writes $\cos^2A$ (and $\sin^2A$, $\tan^2A$) in terms of $\cos2A$ -- trading a squared single angle for an un-squared double angle:

These are the standard route to integrating or simplifying even powers of sine/cosine. Applying the trick twice in a row expresses fourth powers purely in terms of $\cos2x$ and $\cos4x$:

Triple-Angle Identities

Writing $3A=2A+A$ and expanding with the sum formula plus the double-angle identities above builds the triple-angle family.

Identity 3.12. $\sin3A=\sin(2A+A)=\sin2A\cos A+\cos2A\sin A=2\sin A\cos^2A+(1-2\sin^2A)\sin A$. Replacing $\cos^2A$ by $1-\sin^2A$ and collecting terms in $\sin A$:

Identity 3.13. $\cos3A=\cos(2A+A)=\cos2A\cos A-\sin2A\sin A=(2\cos^2A-1)\cos A-2\cos A\sin^2A$. Replacing $\sin^2A$ by $1-\cos^2A$:

Identity 3.14. $\tan3A=\tan(2A+A)=\dfrac{\tan2A+\tan A}{1-\tan2A\tan A}$; substituting $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$ and clearing the compound fraction:

The same "one-step-bigger" idea keeps going: composing the triple- and double-angle identities reaches $\sin4A$ (e.g. $\sin4A=4\sin A\cos^3A-4\cos A\sin^3A$, obtained by factoring $4\sin A\cos A(\cos^2A-\sin^2A)=2(2\sin A\cos A)\cos2A=2\sin2A\cos2A=\sin4A$), and repeating the double-angle sine identity as many times as one likes turns $\sin x$ into $2^{10}\sin\left(\dfrac{x}{2^{10}}\right)$ times a chain of $10$ cosines of successively halved angles -- a pattern that extends to any number of repetitions.

Half-Angle (Sub-multiple) Identities

Every double-angle identity turns into a half-angle identity through the single substitution $2A=\theta$, i.e. $A=\dfrac\theta2$ -- the angle on the right of each formula becomes half of the angle on the left:

Solving the "$1-2\sin^2$" and "$2\cos^2-1$" forms for the half-angle ratio itself gives the two most-used working formulas -- each carrying a $\pm$ sign fixed by which quadrant $\theta/2$ actually lies in:

Worked illustration ($\sin22\tfrac12^\circ$). Take $\theta=45^\circ$ in $\sin\dfrac\theta2=\sqrt{\dfrac{1-\cos\theta}{2}}$ (positive root, since $22\tfrac12^\circ$ is in the first quadrant): with $\cos45^\circ=\dfrac1{\sqrt2}$,

Worked illustration (finding $\sin2\theta$ from $\sin\theta$). If $\sin\theta=\dfrac{12}{13}$ with $\theta$ in the first quadrant, a $5\text{-}12\text{-}13$ right triangle gives $\cos\theta=\dfrac{5}{13}$ (or use $\cos\theta=\sqrt{1-\sin^2\theta}$ directly), so $\sin2\theta=2\sin\theta\cos\theta=2\cdot\dfrac{12}{13}\cdot\dfrac{5}{13}=\dfrac{120}{169}$.

Worked illustration ($\cos2x=\sin x$). Writing $\cos2x=1-2\sin^2x$ turns $\cos2x=\sin x$ into the quadratic $2\sin^2x+\sin x-1=0$, whose roots are $\sin x=\tfrac12$ or $\sin x=-1$. Over $-\pi\le x\le\pi$, $\sin x=\tfrac12$ gives $x=\tfrac\pi6,\tfrac{5\pi}6$ and $\sin x=-1$ gives $x=-\tfrac\pi2$, so the full solution set is $x=-\tfrac\pi2,\tfrac\pi6,\tfrac{5\pi}6$.

Worked illustration -- the golden-ratio exact values ($\sin18^\circ$ and friends). Let $\theta=18^\circ$, so $5\theta=90^\circ$, i.e. $2\theta=90^\circ-3\theta$. Then $\sin2\theta=\sin(90^\circ-3\theta)=\cos3\theta$, i.e. $2\sin\theta\cos\theta=4\cos^3\theta-3\cos\theta$. Since $\cos18^\circ\ne0$, dividing through by $\cos\theta$ gives $2\sin\theta=4\cos^2\theta-3=4(1-\sin^2\theta)-3=1-4\sin^2\theta$, a quadratic in $\sin\theta$:

Since $18^\circ$ is in the first quadrant, $\sin18^\circ>0$, so the $+$ root applies:

$\sin18^\circ=\frac{\sqrt5-1}{4}.$ …