Special Determinants: Trig, Complex, ω, Polynomial

Use $\sin^2\theta + \cos^2\theta = 1$, double-angle, and (for triangle problems) $A + B + C = \pi$. Many such determinants are identically 0 because a trig identity makes rows dependent. Expand only after simplifying with the identity. Triangle reductions you will need: from $A+B+C=\pi$, $\sin(B+C) = \sin(\pi - A) = \sin A$, $\cos(B+C) = -\cos A$, $\tan(B+C) = -\tan A$ — the sine keeps its sign, cosine and tangent flip. And the cosine rule $a^2 = b^2 + c^2 - 2bc\cos A$ collapses a determinant whose rows mix sides and cosines (see the Properties of Triangle notes for the rule itself).

Powers of $i$: $i^1 = i,\ i^2 = -1,\ i^3 = -i,\ i^4 = 1$, repeating every 4. After expanding a complex determinant, write it as $A + iB$ and read off $A$ (real) and $B$ (imaginary), or solve for unknowns by equating real/imaginary parts.

For a non-real cube root of unity $\omega$: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Reduce every power of $\omega$ mod 3, then use the sum relation — a row or column summing to $1 + \omega + \omega^2$ becomes 0, forcing the determinant to 0.

AP/GP rows: three rows in arithmetic progression satisfy $R_1 + R_3 = 2R_2$ (dependent) → determinant 0; GP rows are proportional after a log/ratio step → 0. Determinant as polynomial: expand to a polynomial in $x$ and equate coefficients, or argue the degree to find a specific coefficient. The zero-polynomial argument: a polynomial of degree at most 2 that vanishes at THREE distinct values of $x$ is identically zero — every coefficient is 0. So if a determinant that is quadratic in $x$ is zero at three given points, it is zero for ALL $x$, and any "find the coefficient" question about it answers 0.