Trig Formulas That Win

Why they hold. Rotating the point $(\cos b, \sin b)$ by the angle $a$ lands on $(\cos(a + b), \sin(a + b))$, and in complex numbers that rotation is the product $(\cos a + i\sin a)(\cos b + i\sin b)$; expanding and matching real and imaginary parts gives both addition formulas. Double angles are the case $a = b$, and the tangent formula is the sine formula divided by the cosine formula.

Where it comes from. $\cos 3\theta = \cos(2\theta + \theta) = \cos2\theta\cos\theta - \sin2\theta\sin\theta = (2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta$, and replacing $\sin^2\theta$ by $1 - \cos^2\theta$ leaves $4\cos^3\theta - 3\cos\theta$. So $x = \cos\theta$ turns $4x^3 - 3x = c$ into $\cos 3\theta = c$, which has three solutions spaced $120^\circ$ apart: the cubic's three roots.

Spotting it. Squaring both equations and adding produces $\sin^2 + \cos^2$ twice and the cross terms $2(\cos x\cos y + \sin x\sin y) = 2\cos(x - y)$.

Step 1. $(\sin x + \sin y)^2 + (\cos x + \cos y)^2 = \tfrac14 + \tfrac19 = \tfrac{13}{36}$.

Step 2. The left side is $2 + 2\cos(x - y)$. So $\cos(x - y) = \tfrac12\left(\tfrac{13}{36} - 2\right) = -\tfrac{59}{72}$, and $m + n = 131$.

Spotting it. $8x^3 - 6x = 2(4x^3 - 3x)$, so with $x = \cos\theta$ the equation is $\cos3\theta = \tfrac12$: $3\theta = 60^\circ, 420^\circ, 780^\circ$, i.e. $\theta = 20^\circ, 140^\circ, 260^\circ$. Now it is a Vieta question.

Step 1. For $8x^3 + 0x^2 - 6x - 1$: $e_1 = 0$, $e_2 = -\tfrac68 = -\tfrac34$, $e_3 = \tfrac18$.

Step 2. $\sum\tfrac1r = \tfrac{e_2}{e_3} = \dfrac{-3/4}{1/8} = -6$.

Bonus. $e_1 = 0$ says $\cos20^\circ + \cos140^\circ + \cos260^\circ = 0$: three unit vectors $120^\circ$ apart cancel.

Why it works. Product-to-sum on the last two factors: $\sin(60^\circ - \theta)\sin(60^\circ + \theta) = \tfrac12\left(\cos2\theta - \cos120^\circ\right) = \tfrac12\cos2\theta + \tfrac14 = \tfrac34 - \sin^2\theta$. Multiply by $\sin\theta$: $\tfrac34\sin\theta - \sin^3\theta = \tfrac14(3\sin\theta - 4\sin^3\theta) = \tfrac14\sin3\theta$. The triple-angle formula is what makes the product collapse.

Spotting it. Each angle is double the previous one, so multiplying by $\sin20^\circ$ starts a chain of $2\sin x\cos x = \sin2x$.

Step 1. $8\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ = 4\sin40^\circ\cos40^\circ\cos80^\circ = 2\sin80^\circ\cos80^\circ = \sin160^\circ$.

Step 2. $\sin160^\circ = \sin20^\circ$, so the product is $\tfrac{\sin20^\circ}{8\sin20^\circ} = \tfrac18$.

Spotting it. The angles pair up to $45^\circ$: $3 + 42 = 6 + 39 = \cdots = 45$. For $a + b = 45^\circ$, $\tan(a + b) = 1$ gives $\tan a + \tan b = 1 - \tan a\tan b$, so $(1 + \tan a)(1 + \tan b) = 1 + (\tan a + \tan b) + \tan a\tan b = 2$.

Step 1. The angles $3^\circ, 6^\circ, \dots, 42^\circ$ are $14$ terms, forming $7$ pairs $(3, 42), (6, 39), \dots, (21, 24)$, with no middle term since $22.5$ is not a multiple of $3$.

Step 2. The product is $2^7 = 128$.

Spotting it. $\cos^2 k^\circ + \cos^2(90 - k)^\circ = \cos^2 k^\circ + \sin^2 k^\circ = 1$. Pair from the outside in and handle what is left unpaired.

Step 1. $k = 0$ pairs with $k = 90$: $1 + 0 = 1$. Then $k = 1, \dots, 44$ pair with $89, \dots, 46$: $44$ more pairs, each worth $1$.

Step 2. The middle term $\cos^2 45^\circ = \tfrac12$ is alone.

Step 3. Total $1 + 44 + \tfrac12 = \tfrac{91}{2}$, so $m + n = 93$.

These are what make a long trigonometric sum telescope. Multiplying $\sum\sin(k\theta)$ by $2\sin\tfrac\theta2$ converts every term into a difference of cosines, and the sum collapses to $\dfrac{\sin\tfrac{n\theta}2\sin\tfrac{(n+1)\theta}2}{\sin\tfrac\theta2}$. The trick to remember is the multiplication by $2\sin\tfrac\theta2$, not the final formula.

Sums of $\arctan\dfrac{1}{k^2+k+1}$ telescope because $\dfrac{1}{k^2+k+1} = \dfrac{(k+1)-k}{1+k(k+1)}$, so each term is $\arctan(k+1) - \arctan k$. Recognising a term as a DIFFERENCE is the whole game; the algebraic identity just confirms it.

Step 1 — write $\dfrac{1}{k^2+k+1} = \dfrac{(k+1)-k}{1+k(k+1)}$.

Step 2 — so the term is $\arctan(k+1) - \arctan(k)$.

Step 3 — the sum telescopes to $\arctan(n+1) - \arctan 1 = \arctan(n+1) - \tfrac\pi4$.

Spotting it. Look for $\tfrac{a - b}{1 + ab}$: with $a = k + 1$ and $b = k - 1$, $a - b = 2$ and $1 + ab = k^2$. So each term is a difference of arctangents two apart.

Step 1. $\arctan\dfrac{2}{k^2} = \arctan(k + 1) - \arctan(k - 1)$ (both arctangents are in $[0, \tfrac\pi2)$, so the difference formula applies).

Step 2. Summing to $N$, the terms cancel two apart and leave $\arctan(N + 1) + \arctan N - \arctan1 - \arctan0$.

Step 3. As $N \to \infty$ this tends to $\tfrac\pi2 + \tfrac\pi2 - \tfrac\pi4 - 0 = \tfrac{3\pi}{4}$, so $m + n = 7$.