In some optimization problems, I started getting very odd answers, and have traced it back to solutions of the following equation: $$\cos(t)+\tan(t)\sec(t)=0$$

Mathematica tells me that the solution of this equation (for reals between $-\frac\pi2$ and $\frac\pi2$) is $2 \arctan(a)$ where $a$ a root of the equation $-1 - 2 x + 3 x^2 - 4 x^3 - 3 x^4 - 2 x^5 + x^6$. I'm not super interested in the derivation of this specific solution, but more broadly, how is it that trigonometric equations have solutions in terms of these giant unwieldy polynomials?

Is there a process for deriving said polynomials?


3 Answers 3


Write your equation in the form $$\cos(t)+\frac{\sin(t)}{\cos^2(t)}=0$$ and then make the substitution: $$\sin(t)=\frac{2t}{1+t^2}$$ $$\cos(t)=\frac{1-t^2}{1+t^2}$$ Then you have to solve $${\frac {-{t}^{2}+1}{{t}^{2}+1}}+2\,{\frac {t}{-{t}^{2}+1}}=0$$ for $t$


Any of the six trigonometric functions can be written as a rational function of $\sin{x}$ and $\cos{x}$. If we have a rational function of $\sin{x}$ and $\cos{x}$, it can be written as a ratio of polynomials in $\sin{x}$ and $\cos{x}$. So to solve $f(\sin{x},\cos{x})=0$ for a rational function, it suffices to solve $p(\sin{x},\cos{x})=0$ for the particular polynomial $p$ in the numerator of $f$ (in lowest terms, etc.)

But by the tangent half-angle formulae, if $t=\tan{(x/2)}$, we have $$ \sin{x} = \frac{2t}{1+t^2} , \quad \cos{x} = \frac{1-t^2}{1+t^2}, $$ so the solutions are the solutions of $$ p\left( \frac{2t}{1+t^2} , \frac{1-t^2}{1+t^2} \right) = 0 . $$ Multiplying this by a power of $1+t^2$ then finally gives a polynomial in $t$.

For example, the equation you give is $$ 0 = \cos{t} + \tan{t}\sec{t} = \frac{\cos^2{t}+\sin{t}}{\cos{t}} , $$ so we want to solve $$ 0 = \cos^2{t} + \sin{t} . $$ Using the tangent half-angle formulae, $$ 0 = \frac{(1-t^2)^2}{(1+t^2)^2} + \frac{2t}{1+t^2} = \frac{(1-t^2)^2 + 2t(1+t^2)}{(1+t^2)^2} , $$ and the numerator should simplify into the polynomial you've given.

It's simpler in this case to rewrite $\cos^2{t} = 1-\sin^2{t}$, and then you find that you need to solve $$ 1 + \sin{t} - \sin^2{t} = 0 , $$ which is much easier and won't introduce spurious roots.


Short answer is that trig angular relations are themselves ratios of right triangle length of sides ... like $x = f(\theta)= f(t) $ say.. in the given example.

So if $$ x=\tan \theta= \tan t$$

then $$\cos 2 x = \frac{1-t^2}{1+t^2}$$

etc. and for similar Weierstrass trig relations will define polynomials in a functional polynomial relationship by eliminating of $\theta$ or $t$, leaving behind polynomials.

To be clear the matter of next obtaining roots of the resulting polynomial is entirely a consequential process.


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