Answer to Question #144815 in Trigonometry for saad

Question

Find at least one solution to the following equation:
sin(x^2 − 1)/1 − sin(x^2 − 1) = sin(x) + sin^2(x) + sin^3(x) + sin^4(x) + · · ·

Answer 1

Firstly, let us solve the equation $x^2-1=x$ which is equivalent to $x^2-x-1=0$, and has the solutions $a=\frac{1+\sqrt{5}}{2}$ and $b=\frac{1-\sqrt{5}}{2}$. Therefore, $\sin(a^2-1)=\sin(a)$ and $\sin(b^2-1)=\sin(b)$. Since $\frac{\pi}{2}<a<\pi$ and $-\frac{\pi}{2}<b<0$, we conclude that $|\sin a|<1$ and $|\sin b|<1$. Then the

infinite geometric progression $\sin(a), \sin^2(a), \sin^3(a),..., \sin^4(a), ...$ has a scale factor $\sin a$ and the common ratio $\sin a$ with $|\sin a|<1$. Consequently,

$\sin(a) + \sin^2(a) + \sin^3(a) + \sin^4(a) + · · ·=\frac{\sin(a)}{1-\sin(a)}$.

By analogy, $\sin(b) + \sin^2(b) + \sin^3(b) + \sin^4(b) + · · ·=\frac{\sin(b)}{1-\sin(b)}$.

Therefore, $\frac{\sin(a^2-1)}{1-\sin(a^2-1)}=\frac{\sin(a)}{1-\sin(a)}$ and $\frac{\sin(b^2-1)}{1-\sin(b^2-1)}=\frac{\sin(b)}{1-\sin(b)}$, and we conclude that $a$ and $b$ are the solutions of the equation $\frac{\sin(x^2 − 1)}{1 − \sin(x^2 − 1)} = \sin(x) + \sin^2(x) + \sin^3(x) + \sin^4(x) + · · ·$