Trigonometry & Complex Numbers

Why they hold. The point at angle $\theta$ on the unit circle is $(\cos\theta, \sin\theta)$, and its distance from the origin is $1$: that is the Pythagorean identity. The double-angle formulas are the addition formulas with both angles equal, and $\cos 2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$ follows by replacing $\cos^2$ or $\sin^2$ with the Pythagorean identity. The law of cosines is the distance formula applied to a triangle with one vertex at the origin. Knowing the source tells you which form to reach for: $1 - 2\sin^2\theta$ when you know the sine, $2\cos^2\theta - 1$ when you know the cosine.

Spotting it. Squaring $\sin x + \cos x$ produces $\sin^2 x + \cos^2 x = 1$ plus the cross term $2\sin x\cos x = \sin 2x$. That is the whole trick.

Step 1. $(\sin x + \cos x)^2 = 1 + \sin 2x = \tfrac{49}{25}$, so $\sin 2x = \tfrac{24}{25}$.

Step 2. $(\sin x - \cos x)^2 = 1 - \sin 2x = \tfrac{1}{25}$, so $|\sin x - \cos x| = \tfrac15$.

Step 3. Adding and subtracting, $\sin x$ and $\cos x$ are $\tfrac45$ and $\tfrac35$ in some order. Both orders fit the data, which is exactly why the problem asks for the absolute value.

Spotting it. Three sides and an angle wanted: the law of cosines solved for the cosine.

Step 1. $\cos\theta = \dfrac{5^2 + 8^2 - 7^2}{2 \cdot 5 \cdot 8} = \dfrac{40}{80} = \dfrac12$, so $\theta = 60^\circ$.

Step 2. Area $= \tfrac12 \cdot 5 \cdot 8 \cdot \sin 60^\circ = 10\sqrt3$.

Check with Heron: $s = 10$, and $\sqrt{10 \cdot 5 \cdot 3 \cdot 2} = \sqrt{300} = 10\sqrt3$ ✓

Spotting it. Only $\tan\theta$ is known. Divide the double-angle formulas by $\cos^2\theta = \tfrac{1}{1 + \tan^2\theta}$ to write them in $t = \tan\theta$.

Step 1. $\sin 2\theta = \dfrac{2\sin\theta\cos\theta}{\sin^2\theta + \cos^2\theta} = \dfrac{2t}{1 + t^2} = \dfrac45$.

Step 2. $\cos 2\theta = \dfrac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta} = \dfrac{1 - t^2}{1 + t^2} = -\dfrac35$.

Sense check. $\tan\theta > 1$ means $\theta > 45^\circ$, so $2\theta > 90^\circ$ and a negative cosine is expected.

Why it works. Multiply $r(\cos\alpha + i\sin\alpha)$ by $s(\cos\beta + i\sin\beta)$. The real part is $rs(\cos\alpha\cos\beta - \sin\alpha\sin\beta) = rs\cos(\alpha + \beta)$ and the imaginary part is $rs\sin(\alpha + \beta)$. So lengths multiply and angles add: the addition formulas are hiding inside complex multiplication. Repeating the product $n$ times gives de Moivre.

Why they sum to zero. Let $S = 1 + \omega + \cdots + \omega^{n-1}$ with $\omega = e^{2\pi i/n}$. Multiplying by $\omega$ shifts every term to the next one (and $\omega^n = 1$ wraps the last back to the first), so $\omega S = S$. Since $\omega \ne 1$, $S = 0$. Geometrically: rotating the regular $n$-gon of roots by $\tfrac{360^\circ}{n}$ leaves it unchanged, and the only vector unchanged by a rotation is zero.

Spotting it. A tenth power of a binomial with equal real and imaginary parts: go polar.

Step 1. $|1 + i| = \sqrt2$ and the argument is $45^\circ$.

Step 2. $(1 + i)^{10} = (\sqrt2)^{10}\left(\cos450^\circ + i\sin450^\circ\right) = 32(\cos90^\circ + i\sin90^\circ) = 32i$.

Check a shorter way: $(1 + i)^2 = 2i$, so $(1 + i)^{10} = (2i)^5 = 32i^5 = 32i$ ✓

Spotting it. Going counterclockwise around a square, each side is the previous one turned $90^\circ$ left, and turning by $90^\circ$ is multiplication by $i$.

Step 1. $B - A = 3 + 2i$. Turned left: $i(3 + 2i) = -2 + 3i$.

Step 2. $C = B + (-2 + 3i) = 3 + 6i$ and $D = A + (-2 + 3i) = 4i$.

Check: $D - C = -3 - 2i = -(B - A)$, so opposite sides are parallel and equal ✓

Spotting it. $e^{i\theta} + e^{-i\theta} = 2\cos\theta$, so the condition says $z$ is on the unit circle at angle $\pm20^\circ$.

Step 1. $z^2 - 2\cos20^\circ\,z + 1 = 0$ has roots $z = \cos20^\circ \pm i\sin20^\circ$.

Step 2. By de Moivre, $z^9 = \cos(\pm180^\circ) + i\sin(\pm180^\circ) = -1$ either way, and $\tfrac{1}{z^9} = -1$ too.

Step 3. So $z^9 + z^{-9} = -2$. In general $z^n + z^{-n} = 2\cos(20n)^\circ$.

Powers become multiplication of angles, which is why complex numbers beat trig identities for high powers: $(1+i)^{8} = \left(\sqrt2 e^{i\pi/4}\right)^8 = 16e^{2\pi i} = 16$. Running it backwards gives the $n$-th roots: $z^n = w$ has exactly $n$ solutions, spaced $\tfrac{360^\circ}{n}$ apart on a circle of radius $\sqrt[n]{|w|}$.

The vanishing sum is the engine of the roots-of-unity FILTER: averaging $f(\omega^k)$ over all $k$ keeps exactly the coefficients whose index is divisible by $n$. That is how "how many subsets have size divisible by $3$" becomes a three-term evaluation of $(1+x)^n$ instead of a long sum. Geometrically the roots are the vertices of a regular $n$-gon, which is why they cancel.

Step 1 — polar form: $|1 + i\sqrt3| = 2$ and the argument is $60^\circ$.

Step 2 — De Moivre: $2^6\left(\cos360^\circ + i\sin360^\circ\right) = 64(1 + 0i)$.

Step 3 — so the value is $64$. Expanding the binomial would take six terms and several sign slips; the polar route takes one line.

Spotting it. The $n$ solutions of $z^n = w$ are equally spaced on the circle of radius $\sqrt[n]{|w|}$: a regular $n$-gon.

Step 1. $|-16| = 16$, so the radius is $\sqrt[4]{16} = 2$. The argument of $-16$ is $180^\circ$, so the roots sit at $45^\circ, 135^\circ, 225^\circ, 315^\circ$.

Step 2. Four equally spaced points form a square whose diagonals are diameters of length $4$. Its area is $\tfrac12 \cdot 4 \cdot 4 = 8$.

Spotting it. The answer is $\binom90 + \binom93 + \binom96 + \binom99$, every third coefficient of $(1 + x)^9$. Averaging $(1 + x)^9$ over the cube roots of unity keeps exactly those terms, because $1 + \omega^k + \omega^{2k}$ is $3$ when $3 \mid k$ and $0$ otherwise.

Step 1. $\tfrac13\left[(1 + 1)^9 + (1 + \omega)^9 + (1 + \omega^2)^9\right]$ is the count.

Step 2. $1 + \omega = -\omega^2$, so $(1 + \omega)^9 = -\omega^{18} = -1$; likewise $(1 + \omega^2)^9 = -\omega^9 = -1$.

Step 3. The count is $\tfrac13(512 - 1 - 1) = 170$. Check: $1 + 84 + 84 + 1 = 170$ ✓