Advanced Trigonometry Compendium

$\sin\theta\csc\theta = 1, \qquad \cos\theta\sec\theta = 1, \qquad \tan\theta\cot\theta = 1$

$\tan\theta = \dfrac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \dfrac{\cos\theta}{\sin\theta}$

$\sin^2\theta + \cos^2\theta = 1 \qquad 1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \csc^2\theta$

• Period of $f(ax + b)$ is $T/|a|$ if the period of $f(x)$ is $T$.
• Period of $|\sin x|, |\cos x|, |\tan x|, |\cot x|, |\sec x|, |\csc x|$ is $\pi$.

$\sin(-\theta) = -\sin\theta, \qquad \cos(-\theta) = \cos\theta, \qquad \tan(-\theta) = -\tan\theta$

$1 + \sin 2\theta = (\sin\theta + \cos\theta)^2$

$1 - \sin 2\theta = (\sin\theta - \cos\theta)^2 = (\cos\theta - \sin\theta)^2$

$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$

$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$

$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$

• $\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B = \cos^2 B - \cos^2 A$
• $\cos(A+B)\cos(A-B) = \cos^2 A - \sin^2 B = \cos^2 B - \sin^2 A$
• If $\theta_1 + \theta_2 = \pi/4$, then $(1 + \tan\theta_1)(1 + \tan\theta_2) = 2$.
• $\tan(45° + \alpha) - \tan(45° - \alpha) = 2\tan 2\alpha$

$2\sin A\cos B = \sin(A+B) + \sin(A-B)$

$2\cos A\sin B = \sin(A+B) - \sin(A-B)$

$2\cos A\cos B = \cos(A+B) + \cos(A-B)$

$2\sin A\sin B = \cos(A-B) - \cos(A+B)$

$\sin C + \sin D = 2\sin\!\left(\frac{C+D}{2}\right)\cos\!\left(\frac{C-D}{2}\right)$

$\sin C - \sin D = 2\cos\!\left(\frac{C+D}{2}\right)\sin\!\left(\frac{C-D}{2}\right)$

$\cos C + \cos D = 2\cos\!\left(\frac{C+D}{2}\right)\cos\!\left(\frac{C-D}{2}\right)$

$\cos C - \cos D = -2\sin\!\left(\frac{C+D}{2}\right)\sin\!\left(\frac{C-D}{2}\right)$

$\sin 2A = 2\sin A\cos A = \frac{2\tan A}{1 + \tan^2 A}$

$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A = \frac{1 - \tan^2 A}{1 + \tan^2 A}$

$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$

• $1 - \cos 2A = 2\sin^2 A$
• $1 + \cos 2A = 2\cos^2 A$
• $\dfrac{\sin 2A}{1 + \cos 2A} = \tan A$
• $\cot A - \tan A = 2\cot 2A$
• $\tan A + \cot A = 2\csc 2A$
• $\csc\theta + \cot\theta = \cot(\theta/2)$

$\sin 3A = 3\sin A - 4\sin^3 A$

$\cos 3A = 4\cos^3 A - 3\cos A$

$\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$

$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$

$\tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}$

$\sin^2 A = \frac{1 - \cos 2A}{2} \qquad \cos^2 A = \frac{1 + \cos 2A}{2}$

$\sin^3 x = \frac{3\sin x - \sin 3x}{4} \qquad \cos^3 x = \frac{3\cos x + \cos 3x}{4}$

$\sin^4 x = \frac{3 - 4\cos 2x + \cos 4x}{8} \qquad \cos^4 x = \frac{3 + 4\cos 2x + \cos 4x}{8}$

If $\dfrac{\sin^4 x}{a} + \dfrac{\cos^4 x}{b} = \dfrac{1}{a+b}$, then $\dfrac{\sin^{2N} x}{a^{N-1}} + \dfrac{\cos^{2N} x}{b^{N-1}} = \dfrac{1}{(a+b)^{N-1}}$ for integer $N \ge 2$.

$\sin(\theta/2) = \pm\sqrt{\frac{1 - \cos\theta}{2}} \qquad \cos(\theta/2) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$

$\tan(\theta/2) = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}$

Sign depends on the quadrant of $\theta/2$.

• $\sin\theta\,\sin(60°-\theta)\sin(60°+\theta) = \tfrac{1}{4}\sin 3\theta$
• $\cos\theta\,\cos(60°-\theta)\cos(60°+\theta) = \tfrac{1}{4}\cos 3\theta$
• $\tan\theta\,\tan(60°-\theta)\tan(60°+\theta) = \tan 3\theta$
• $\cos^2\theta + \cos^2(60°+\theta) + \cos^2(60°-\theta) = \tfrac{3}{2}$
• $\sin^2\theta + \sin^2(60°+\theta) + \sin^2(60°-\theta) = \tfrac{3}{2}$
• $\cos^3\theta + \cos^3(120°+\theta) + \cos^3(240°+\theta) = \tfrac{3}{4}\cos 3\theta$
• $\sin^3\theta + \sin^3(120°+\theta) + \sin^3(240°+\theta) = -\tfrac{3}{4}\sin 3\theta$
• $\tan\theta + \tan(60°+\theta) - \tan(60°-\theta) = 3\tan 3\theta$
• $\tan^2\theta + \tan^2(60°-\theta) + \tan^2(60°+\theta) = 6 + 9\tan^2 3\theta$

$\sin(\theta_1 + \cdots + \theta_n) = (\cos\theta_1 \cdots \cos\theta_n)(S_1 - S_3 + S_5 - \cdots)$

$\cos(\theta_1 + \cdots + \theta_n) = (\cos\theta_1 \cdots \cos\theta_n)(1 - S_2 + S_4 - \cdots)$

$\tan(\theta_1 + \cdots + \theta_n) = \frac{S_1 - S_3 + S_5 - \cdots}{1 - S_2 + S_4 - \cdots}$

For $t = \tan\theta$:

$\tan(n\theta) = \frac{\binom{n}{1}t - \binom{n}{3}t^3 + \cdots}{1 - \binom{n}{2}t^2 + \cdots}$

If $\alpha \ne 2m\pi$:

$\sum_{k=0}^{n-1}\sin(\theta + k\alpha) = \frac{\sin\!\left(\theta + \frac{(n-1)\alpha}{2}\right)\sin(n\alpha/2)}{\sin(\alpha/2)}$

$\sum_{k=0}^{n-1}\cos(\theta + k\alpha) = \frac{\cos\!\left(\theta + \frac{(n-1)\alpha}{2}\right)\sin(n\alpha/2)}{\sin(\alpha/2)}$

• $\displaystyle\sum_{k=1}^{n} k\sin(kx) = \frac{(n+1)\sin(nx) - n\sin((n+1)x)}{4\sin^2(x/2)}$
• $\displaystyle\sum_{k=1}^{n} k\cos(kx) = \frac{(n+1)\cos(nx) - n\cos((n+1)x) - 1}{4\sin^2(x/2)}$
• $\displaystyle\sum_{k=1}^{n} \sin((2k-1)x) = \frac{\sin^2(nx)}{\sin x}$
• $\displaystyle\sum_{k=1}^{n} \cos((2k-1)x) = \frac{\sin(2nx)}{2\sin x}$
• $\displaystyle\sum_{k=1}^{n-1} k\cos(2k\pi/n) = -n/2$ for $n \ge 2$

• $\displaystyle\sum_{j=1}^{N}\sin^2(jx) = \frac{N}{2} - \frac{\sin(Nx)\cos((N+1)x)}{2\sin x}$
• $\displaystyle\sum_{j=1}^{N}\cos^2(jx) = \frac{N}{2} + \frac{\sin(Nx)\cos((N+1)x)}{2\sin x}$
• $\displaystyle\sum_{k=1}^{n}\frac{1}{\sin 2^k x} = \cot x - \cot 2^n x$

• $\displaystyle\sum_{k=0}^{n-1}\cot\!\left(x + \frac{k\pi}{n}\right) = n\cot(nx)$
• $\displaystyle\sum_{k=0}^{n-1}\csc^2\!\left(x + \frac{k\pi}{n}\right) = n^2\csc^2(nx)$

$\prod_{k=0}^{n-1}\cos(2^k x) = \frac{\sin(2^n x)}{2^n \sin x} \qquad (\sin x \ne 0)$

$\prod_{k=1}^{n}\cos\!\left(\frac{\theta}{2^k}\right) = \frac{\sin\theta}{2^n\sin(\theta/2^n)} \qquad (\sin(\theta/2^n) \ne 0)$

• Morrie’s Law: $\cos 20°\cos 40°\cos 80° = 1/8$
• $\cos 6°\cos 42°\cos 66°\cos 78° = 1/16$
• $\sin 5°\sin 15°\sin 25° \cdots \sin 85° = \dfrac{\sqrt{2}}{256}$
• $\displaystyle\prod_{k=1}^{89}\tan(k°) = 1$

For $n \in \mathbb{N}$, $n \ge 2$:

• $\displaystyle\prod_{k=1}^{n-1}\sin\!\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$
• $\displaystyle\prod_{k=1}^{n-1}\cos\!\left(\frac{k\pi}{n}\right) = \begin{cases} 0 & n \text{ even} \\ 1/2^{n-1} & n \text{ odd}, n > 1 \end{cases}$
• $\displaystyle\prod_{k=1}^{n}\sin\!\left(\frac{(2k-1)\pi}{2n}\right) = \frac{1}{2^{n-1}}$
• $\displaystyle\prod_{k=1}^{n}\sin\!\left(\frac{k\pi}{2n+1}\right) = \frac{\sqrt{2n+1}}{2^n}$
• $\displaystyle\prod_{k=1}^{n-1}\csc\!\left(\frac{k\pi}{n}\right) = \frac{2^{n-1}}{n}$
• $\displaystyle\prod_{k=1}^{n-1}(1 - \cos(2k\pi/n)) = n^2/2^{n-1}$

• $\displaystyle\sum_{k=0}^{n-1}\csc(2^k A) = \cot(A/2) - \cot(2^{n-1}A)$
• $\displaystyle\sum_{k=0}^{n-1}2^k\tan(2^k\theta) = \cot\theta - 2^n\cot(2^n\theta)$
• $\displaystyle\sum_{r=1}^{n}\frac{\sin\alpha}{\cos r\alpha\cos(r+1)\alpha} = \tan((n+1)\alpha) - \tan\alpha$
• $\displaystyle\sum_{k=1}^{n}\frac{1}{\sin(k\alpha)\sin((k+1)\alpha)} = \frac{1}{\sin\alpha}\big(\cot\alpha - \cot((n+1)\alpha)\big)$

$\sum_{k=0}^{n}\binom{n}{k}\sin(kx) = 2^n\cos^n(x/2)\sin(nx/2)$

$\sum_{k=0}^{n}\binom{n}{k}\cos(kx) = 2^n\cos^n(x/2)\cos(nx/2)$

With alternating signs:

$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\cos(kx) = 2^n\sin^n(x/2)\cos\!\left(\frac{n(x-\pi)}{2}\right)$

$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\sin(kx) = 2^n\sin^n(x/2)\sin\!\left(\frac{n(x-\pi)}{2}\right)$

• $\displaystyle\sum_{j=1}^{\infty}\frac{\sin jx}{j} = \frac{\pi - x}{2}$ for $x \in (0, 2\pi)$
• $\displaystyle\sum_{j=1}^{\infty}\frac{\cos jx}{j} = -\ln|2\sin(x/2)|$ for $x \in (0, 2\pi)$
• $\displaystyle\sum_{j=1}^{\infty}\frac{\cos jx}{j^2} = \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6}$ for $x \in [0, 2\pi]$
• $\displaystyle\sum_{j=1}^{\infty}(-1)^{j-1}\frac{\sin jx}{j} = \frac{x}{2}$ for $x \in (-\pi, \pi)$
• $\displaystyle\sum_{j=1}^{\infty}(-1)^{j-1}\frac{\cos jx}{j} = \ln|2\cos(x/2)|$ for $x \in (-\pi, \pi)$
• $\displaystyle\sum_{n=1}^{\infty}\frac{\sin^2(nx)}{n^2} = \frac{\pi x - x^2}{2}$ for $x \in [0, \pi]$

$\theta = \tan\theta - \dfrac{\tan^3\theta}{3} + \dfrac{\tan^5\theta}{5} - \cdots$ for $|\tan\theta| \le 1$, $-\pi/4 \le \theta \le \pi/4$.

For $|r| < 1$:

$\sum_{j=1}^{\infty}r^j\sin(j\theta) = \frac{r\sin\theta}{1 - 2r\cos\theta + r^2} \qquad \sum_{j=0}^{\infty}r^j\cos(j\theta) = \frac{1 - r\cos\theta}{1 - 2r\cos\theta + r^2}$

For $z \notin \mathbb{Z}$:

$\pi\cot(\pi z) = \frac{1}{z} + 2z\sum_{j=1}^{\infty}\frac{1}{z^2 - j^2} \qquad \frac{\pi}{\sin(\pi z)} = \frac{1}{z} + 2z\sum_{j=1}^{\infty}\frac{(-1)^j}{z^2 - j^2}$

$\pi\tan(\pi z) = 2z\sum_{j=0}^{\infty}\frac{1}{(j + 1/2)^2 - z^2}$

$\operatorname{Cl}_2(x) = \displaystyle\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^2}$. $\quad \operatorname{Cl}_2(\pi/2) = G$ (Catalan’s constant).

$\operatorname{Cl}_3(x) = \displaystyle\sum_{k=1}^{\infty}\frac{\cos(kx)}{k^3}$.

• If $\theta = 2\pi/7$: $\tan\theta\tan 2\theta + \tan 2\theta\tan 4\theta + \tan 4\theta\tan\theta = -7$
• $\displaystyle\sum_{k=1}^{n-1}(n-k)\cos(2k\pi/n) = -n/2$ for $n \ge 3$
• $1 + \displaystyle\sum_{k=1}^{n-1}\cos(2k\pi/n) = 0$ for $n > 1$
• $\displaystyle\sum_{k=0}^{\infty}\frac{1}{2^k}\tan\!\left(\frac{x}{2^k}\right) = \frac{1}{x} - \cot x$ $(x \ne m\pi)$

• $\displaystyle\prod_{j=1}^{\infty}\!\left(1 + \frac{1}{j^2}\right) = \frac{\sinh\pi}{\pi}$
• $\displaystyle\prod_{j=1}^{\infty}(1 + x^{2^j}) = \frac{1}{1-x}$ for $|x| < 1$

• $\displaystyle\prod_{r=1}^{n}\tan\!\left(\frac{r\pi}{2n+1}\right) = \sqrt{2n+1}$
• $\displaystyle\prod_{r=1}^{n}\cos\!\left(\frac{r\pi}{2n+1}\right) = \frac{1}{2^n}$
• $\displaystyle\sum_{r=1}^{n}\tan^2\!\left(\frac{r\pi}{2n+1}\right) = n(2n+1)$
• $\displaystyle\sum_{k=1}^{n-1}\sin^{2m}\!\left(\frac{k\pi}{n}\right) = \frac{n}{2^{2m}}\binom{2m}{m}$ for $2m < n$

To sum $\sum a_k\cos(k\theta)$ or $\sum a_k\sin(k\theta)$: form $S_c + iS_s = \sum a_k e^{ik\theta}$. Sum the complex series, then separate $S_c = \operatorname{Re}(\cdot)$ and $S_s = \operatorname{Im}(\cdot)$.

• Sum of roots: $\displaystyle\sum_{k=0}^{n-1}z_k = 0$ for $n > 1$
• Product of roots: $\displaystyle\prod_{k=0}^{n-1}z_k = (-1)^{n-1}$
• $x^n - 1 = \displaystyle\prod_{k=0}^{n-1}(x - e^{i2k\pi/n})$
• Roots of $x^n + 1 = 0$: $e^{i(2k+1)\pi/n}$ for $k = 0, 1, \ldots, n-1$
• $\displaystyle\sum_{k=0}^{n-1}\cos\!\left(\frac{(2k+1)\pi}{n}\right) = 0$

Roots of $T_n(x)$: $x_j = \cos\!\left(\frac{(2j-1)\pi}{2n}\right)$ for $j = 1, \ldots, n$.

Roots of $U_{n-1}(x)$: $x_j = \cos\!\left(\frac{j\pi}{n}\right)$ for $j = 1, \ldots, n-1$.

Even $n$: $\cos^n\theta = \dfrac{1}{2^{n-1}}\!\left[\displaystyle\sum_{k=0}^{n/2-1}\binom{n}{k}\cos((n-2k)\theta) + \frac{1}{2}\binom{n}{n/2}\right]$

Odd $n$: $\cos^n\theta = \dfrac{1}{2^{n-1}}\displaystyle\sum_{k=0}^{(n-1)/2}\binom{n}{k}\cos((n-2k)\theta)$

Even $n$: $\sin^n\theta = \dfrac{(-1)^{n/2}}{2^{n-1}}\!\left[\displaystyle\sum_{k=0}^{n/2-1}(-1)^k\binom{n}{k}\cos((n-2k)\theta) + \frac{(-1)^{n/2}}{2}\binom{n}{n/2}\right]$

Odd $n$: $\sin^n\theta = \dfrac{(-1)^{(n-1)/2}}{2^{n-1}}\displaystyle\sum_{k=0}^{(n-1)/2}(-1)^k\binom{n}{k}\sin((n-2k)\theta)$

• $\tan 70° = \tan 20° + 2\tan 50°$
• $\tan 50° + \tan 60° + \tan 70° = \tan 80°$
• $\tan^2(π/16) + \tan^2(2π/16) + \cdots + \tan^2(7π/16) = 35$
• $(1 + \tan 1°)(1 + \tan 2°) \cdots (1 + \tan 44°) = 2^{22}$
• $\sin^2 18° + \sin^2 30° = \sin^2 36°$
• $\tan^2 36°\cdot\tan^2 72° = 5$
• $\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac{1}{2}$
• $4\sin\frac{2\pi}{7} - \tan\frac{\pi}{7} = \sqrt{7}$

• Roots of $8x^3 - 4x^2 - 4x + 1 = 0$ are $\cos(\pi/7), \cos(3\pi/7), \cos(5\pi/7)$
• $\displaystyle\sum_{k=1}^{n}\cot^2\!\left(\frac{k\pi}{2n+1}\right) = \frac{n(2n-1)}{3}$
• $\displaystyle\sum_{k=1}^{n-1}\tan^2\!\left(\frac{k\pi}{2n}\right) = \frac{(n-1)(2n-1)}{3}$
• $\displaystyle\sum_{k=1}^{n-1}\csc^2\!\left(\frac{k\pi}{2n}\right) = \frac{2(n^2-1)}{3}$
• For $n$ odd: $\displaystyle\sum_{k=1}^{(n-1)/2}\sec^2\!\left(\frac{k\pi}{n}\right) = \frac{n^2-1}{4}$
• $\tan^2 10° + \tan^2 50° + \tan^2 70° = 9$

• $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
• $\cot(A/2) + \cot(B/2) + \cot(C/2) = \cot(A/2)\cot(B/2)\cot(C/2)$
• $\displaystyle\sum_{\text{cyc}}\frac{\cos A}{\sin B\sin C} = 2$

• If $A + B + C = n\pi$: $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
• If $A + B + C = (2n+1)\pi/2$: $\tan A\tan B + \tan B\tan C + \tan C\tan A = 1$
• If $\cos(\alpha-\beta) + \cos(\beta-\gamma) + \cos(\gamma-\alpha) = -3/2$: $\sum\cos\alpha = 0$ and $\sum\sin\alpha = 0$
• $\tan(x-y) + \tan(y-z) + \tan(z-x) = \tan(x-y)\tan(y-z)\tan(z-x)$
• If $A + B = 45°$: $(1 + \tan A)(1 + \tan B) = 2$