# Does an extraneous root of a polynomial in trigonometric functions have meaning? (Eg, the root $\cos\theta=-2$ of $\cos^2\theta+3\cos\theta+2=0$.)

Does an extraneous root of a polynomial in trigonometric functions have meaning?

E.g., $$\cos^2{\theta} + 3\cos{\theta} + 2 = 0$$ has two roots, $$\cos{\theta} = -1$$ and $$\cos{\theta} = -2$$. The second root is usually discarded as it is beyond the range of the cosine function. But does it have any geometric, analytical, etc, meaning?

• One may write $\cos t= -2$ as $e^{it}+e^{-it}+4=0 \implies e^{2it}+4e^{it}+1=0$ and get $t$ in the complex domain after solving the quadratic. Nov 4, 2020 at 12:25
• Why am I being downvoted? Nov 4, 2020 at 12:26
• @Arvind Suresh Some people don't like complex roots at all! Nov 4, 2020 at 14:22
• By complex I mean non-real. Nov 4, 2020 at 14:40

Geometrically, discarding $$\cos\theta =-2$$ is to discard those roots lying outside the the real axis of the complex plane, i.e. restricting the domain of $$\theta$$ to real values.

If, instead, the domain of $$\theta$$ is the full complex space, you need to solve for the complex roots, as follows

$$2=-\cos\theta= \cos(2\pi n + \pi-\theta) = \cosh[-i (2\pi n + \pi-\theta)]$$

which yields

$$\theta = (2n+1)\pi \pm i \cosh^{-1}(2)$$

You may write $$\cos t= -2$$ as $$\frac{e^{it}+e^{-it}}{2}=-2 \implies 4^{it}+e^{-it}+4=0 \implies e^{2it}+4e^{it}+1=0 \implies e^{it}= \frac{-4\pm\sqrt{12}}{2}$$ $$=-2\pm\sqrt{3}\implies t=-i \ln (-2\pm \sqrt{3}).$$ Using $$(-1)=e^{2n+1)i\pi}$$, we can write two branches of solutions as: $$t_1=(2n+1)\pi+i \ln(2-\sqrt{3}). ~~t_2=(2n+1)\pi+i\ln (2+\sqrt{3}), n=0,1,2...$$ Or alternatively, two complex conjugate roots as the original equation is real. $$t=(2n+1)\pi\pm i\ln (2+\sqrt{3}), n=0,1,2,...$$ ^n=0\$ guves the principal branch.

• there's a small typo in your last line Nov 4, 2020 at 20:33