The Acute Obtuse Combination

Since $90^\circ < x < 180^\circ$ , let $x = 90^\circ + \theta$ , so that both $0 < \theta, y < 90^\circ$ . Then we have:

$\begin{aligned} \sin x & = 2 \sin (x + 2y) & \small \blue{\text{Let }x = 90^\circ + \theta} \\ \sin (90^\circ + \theta) & = 2 \sin (90^\circ + \theta + 2y) & \small \blue{\text{As }\sin (180^\circ - \phi) = \sin \phi} \\ \sin (90^\circ - \theta) & = 2 \sin (90^\circ - \theta - 2y) & \small \blue{\text{and }\sin (90^\circ - \phi) = \cos \phi} \\ \cos \theta & = 2 \cos (\theta + 2y) \\ & = 2\cos \theta \cos 2y - 2\sin \theta \sin 2y \\ 1 & = 2 \cos 2y - 2 \tan \theta \sin 2y \end{aligned}$

Using half-angle tangent substitution and let $t = \tan y$ .

$\begin{aligned} \frac {2(1-t^2)}{1+t^2} - \frac {4t\tan \theta}{1+t^2} & = 1 \\ \implies 3t^2 + 4t \tan \theta - 1 & = 0 \\ \implies t = \tan y & = \frac {\sqrt{4\tan^2 \theta + 3}-2\tan \theta}3 \end{aligned}$

Note that $\tan y$ decreases with $\theta$ monotonically. Therefore,

$\begin{aligned} \lim_{\theta \to 90^\circ} \frac {\sqrt{4\tan^2 \theta + 3}-2\tan \theta}3 < & \tan y < \lim_{\theta \to 0^\circ} \frac {\sqrt{4\tan^2 \theta + 3}-2\tan \theta}3 \\ 0 < & \tan y < \frac 1{\sqrt 3} \end{aligned}$

Hence $\boxed {0 < \tan y < \frac 1{\sqrt 2}}$ .

Expanding the right hand side of the given equation and simplifying we get $\tan x=\dfrac{2\sin (2y)}{1-2\cos (2y)}$ . Since $x$ is obtuse, $\tan x<0$ . Since $y$ is acute, $\sin (2y)>0$ . Therefore $1-2\cos (2y)<0$ , or $\cos (2y)>\dfrac{1}{2}$ or $y<\dfrac{π}{6}$ . This implies $\tan y<\dfrac{1}{√3}<\dfrac{1}{√2}$ . Since $y$ is acute, therefore $\tan y>0$ , and hence $\boxed {0<\tan y<\dfrac{1}{√2}}$ .