Triple-angle identities: remember, derive, and apply them

The three identities

$$\boxed{\cos3\theta=4\cos^3\theta-3\cos\theta,}$$ $$\boxed{\sin3\theta=3\sin\theta-4\sin^3\theta,}$$ $$\boxed{\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.}$$

The first two hold for every real $\theta$. The tangent formula requires that both tangents be defined; in particular $\cos\theta\ne0$ and $1-3\tan^2\theta\ne0$.

The original Japanese article uses a sound-based mnemonic for cosine. In English, a useful structural reminder is “cosine: four cubes minus three singles; sine: three singles minus four cubes.” Equivalently, swap cosine for sine in the cosine polynomial and reverse its sign.

Quick checks for a remembered formula

For cosine, test $\theta=\pi/6$. Since $\cos3\theta=0$, the right side should give

$$4\left(\frac{\sqrt3}{2}\right)^3-3\frac{\sqrt3}{2}=0.$$

Swapping the coefficients would instead give $-7\sqrt3/8$. However, negating the whole correct formula would still give zero at this test angle. Also test $\theta=0$: the answer must be $4-3=1$, not $-1$.

For sine, test $\theta=\pi/3$ to obtain zero, then $\theta=\pi/2$ to obtain $3-4=-1=\sin(3\pi/2)$. These tests detect common memory slips. Agreement at finitely many angles is not a proof of an identity.

Derive the formulas from angle addition

For cosine,

$$\begin{aligned} \cos3\theta &=\cos2\theta\cos\theta-\sin2\theta\sin\theta\\ &=(2\cos^2\theta-1)\cos\theta -2\sin^2\theta\cos\theta\\ &=4\cos^3\theta-3\cos\theta. \end{aligned}$$

For sine,

$$\begin{aligned} \sin3\theta &=\sin2\theta\cos\theta+\cos2\theta\sin\theta\\ &=2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta\\ &=3\sin\theta-4\sin^3\theta. \end{aligned}$$

To obtain tangent, divide the two expressions. Setting $t=\tan\theta$ and dividing the cubic expressions by $\cos^3\theta$ gives

$$\tan3\theta=\frac{3t(1+t^2)-4t^3}{4-3(1+t^2)} =\frac{3t-t^3}{1-3t^2}.$$

You can also apply tangent addition to $2\theta+\theta$ and substitute $\tan2\theta=2t/(1-t^2)$. That intermediate derivation excludes $t=\pm1$, where $\tan2\theta$ is undefined; the sine/cosine derivation verifies the final formula there whenever both sides of the final identity are defined.

Example 1: turn a trigonometric expression into a polynomial

The source uses the first problem from the 2017 University of Tokyo science-track paper. The intended quotient, consistent with its polynomial solution, is

$$f(\theta)=\cos3\theta+a\cos2\theta+b\cos\theta, \qquad g(\theta)=\frac{f(\theta)-f(0)}{\cos\theta-1},\quad0<\theta<\pi.$$

The source screenshot misplaces the $-1$ outside the denominator. Here it belongs inside the denominator; the two expressions are not interchangeable.

Express $f,g$ as polynomials in $x=\cos\theta$, then determine when $g$ has attained minimum zero on the given interval.

The double- and triple-angle identities give

$$f=4x^3+2ax^2+(b-3)x-a.$$

Since $f(0)$ in the trigonometric notation is the value at $\theta=0$, corresponding to $x=1$, division by $x-1$ yields

$$g=4x^2+2(a+2)x+2a+b+1.$$

Completing the square,

$$g=4\left(x+\frac{a+2}{4}\right)^2+b+1-\frac{(a-2)^2}{4}.$$

The range of $x$ is the open interval $(-1,1)$. The quadratic attains its minimum there exactly when its vertex lies inside that interval. Requiring the minimum to be zero gives

$$\boxed{-6<a<2,\qquad b=\frac{(a-2)^2}{4}-1.}$$

The parameter locus is that open arc of the upward parabola in the $(a,b)$-plane, excluding both endpoint values of $a$. A vertex at an excluded endpoint gives an infimum, not an attained minimum.

Example 2: a cubic equation and powers of a complex number

Let $F(x)=8x^3-6x+1$.

1. Show that $F(\cos(2\pi/9))=0$.
2. Find its three distinct real roots and identify the largest.
3. Prove $\cos(\pi/4)<3/4<\cos(2\pi/9)$.
4. For $z=3/4+i\sqrt7/4$, find the smallest positive integer $n$ such that $z^n$ has positive real part and negative imaginary part, while $z^{n+1}$ has both parts positive.

The three roots

The triple-angle identity gives

$$F(\cos\phi)=2\cos3\phi+1.$$

Thus $\cos(2\pi/9)$, $\cos(4\pi/9)$, and $\cos(8\pi/9)$ are roots. They are distinct because cosine strictly decreases on $(0,\pi)$, and a cubic has at most three distinct roots. Their decreasing order is the order just listed, so $\cos(2\pi/9)$ is the largest.

Locate the largest root

First $\cos(\pi/4)=\sqrt2/2<3/4$, as follows by squaring positive quantities. Also

$$F'(x)=24x^2-6>0\quad(x>1/2),\qquad F(3/4)=-\frac18<0.$$

Since $2\pi/9<\pi/4$, the largest root is greater than $\sqrt2/2>1/2$. Strict increase on this interval and $F(\cos(2\pi/9))=0$ imply $3/4<\cos(2\pi/9)$.

Track the argument of the complex powers

We have $|z|=1$ and both parts positive. Write $z=\cos\theta+i\sin\theta$. The cosine bounds give

$$\frac{2\pi}{9}<\theta<\frac\pi4.$$

By de Moivre's formula, $z^k=\cos(k\theta)+i\sin(k\theta)$. These strict angle ranges locate the powers:

| $k$ | Interval for $k\theta$ | Quadrant of $z^k$ | | --- | --- | --- | | 1 | $(2\pi/9,\pi/4)$ | I | | 2 | $(4\pi/9,\pi/2)$ | I | | 3 | $(2\pi/3,3\pi/4)$ | II | | 4 | $(8\pi/9,\pi)$ | II | | 5 | $(10\pi/9,5\pi/4)$ | III | | 6 | $(4\pi/3,3\pi/2)$ | III | | 7 | $(14\pi/9,7\pi/4)$ | IV | | 8 | $(16\pi/9,2\pi)$ | IV | | 9 | $(2\pi,9\pi/4)$ | I |

Only $k=7,8$ among the first eight powers have the required signs for $z^k$. The next power after $k=7$ is still in quadrant IV; the next after $k=8$ is in quadrant I. Hence

$$\boxed{n=8.}$$

The condition concerns two consecutive powers. Showing merely that one power lies in quadrant IV would give an incomplete answer.

Example 3: a swept solid and a sine-cubed integral

A segment $AB$ of length two moves in space with $A$ in the plane $z=0$ and $C=(0,0,1)$ lying on the segment. Let $K$ be the region it can sweep. Find the volume of the part of $K$ with $z\ge1$.

Write the distance of $A$ from the vertical axis as $r=\cot\theta$, where $\pi/6\le\theta\le\pi/2$. Then $AC=\csc\theta\le2$. Above $C$, the other endpoint $B$ has radial coordinate and height

$$R=(2-\csc\theta)\cos\theta=(2\sin\theta-1)\cot\theta, \qquad z=2\sin\theta.$$

At a fixed height $z\in[1,2]$, the largest attainable radius is therefore

$$R(z)=(z-1)\frac{\sqrt{4-z^2}}{z}.$$

To justify the filled section, any point at radius $\rho$ and height $z>1$ lies on a ray from $C$ that reaches $z=0$ at radius $\rho/(z-1)$. The length from that point to this plane is $z\sqrt{1+(\rho/(z-1))^2}$. It fits on a segment of length two exactly when $\rho\le R(z)$. Thus the section is the full disk, not just its boundary.

Consequently,

$$\begin{aligned} V&=\pi\int_1^2\frac{(z-1)^2(4-z^2)}{z^2}\,dz\\ &=\pi\int_1^2\left(-z^2+2z+3-\frac8z+\frac4{z^2}\right)\,dz\\ &=\boxed{\pi\left(\frac{17}{3}-8\log2\right)}. \end{aligned}$$

An angle substitution $z=2\sin\theta$ gives the same integral. In that approach the cubic sine term can be reduced using

$$4\sin^3\theta=3\sin\theta-\sin3\theta.$$

Alternatively, reserve one sine factor and use $\sin^2\theta=1-\cos^2\theta$:

$$\int\sin^3\theta\,d\theta =-\cos\theta+\frac13\cos^3\theta+C.$$

The triple-angle identity is a useful simplification, rather than the only way to integrate an odd trigonometric power.