I was playing with math, and then I tried to rewrite some cubic equation with sine power reduction formula $$y^3 + my^2 + ny + d = 0.$$ Let $$y = \sin(x).$$ Then $$y^2 = \frac{1 - \cos(2x)}{2},$$ $$y^3 = \frac{3\sin(x) - \sin(3x)}{4}.$$ So the equation will be
$$\frac{3\sin(x)-\sin(3x)}{4} + m\frac{1 - \cos(2x)}{2} + n\cdot\sin(x) +d.$$ We can redefine constant multipliers and we will end with $$a\cdot\sin(x) + b\cdot\cos(2x) + c\cdot\sin(3x) +L = 0.$$ What would be the best approach to solve this without needing to solve polynomial equations whose degree is greater or equal to 3?


1 Answer 1


This approach has been used to resolve the diminished cubic $x^3 = 3px - 2q$

$x = 2\sqrt p\cos\theta\\ 8p\sqrt p\cos^3 \theta = 6p\sqrt p \cos\theta + 2q\\ 2p\sqrt p(4\cos^3 \theta - 3\cos\theta) = 2q\\ \cos 3\theta = \frac {q}{p\sqrt p}$

We detour into the complex plane if the original cubic does not have 3 real roots.

The substitution $y = x - \frac {m}{3}$ will allow you to turn your initial cubic into a diminished cubic.


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