I'm doing this calculation whereby I determine where the light source would be reflected on the sphere to the eye relative to its position. Assuming light travels in a straight line and the ray is perpendicular to the eye, I would like to find where it would reflect to the eye on the surface normal of the sphere.
Plan view of the sphere diagram
The radius of the sphere is $1$, the $x$-distance from the eye to the center of the sphere is $\frac{1}{\sin15}$ and $y$-distance is $1$. With that, I identified the triangle where I can calculate the angle of incidence using sine rule. With the light ray being perpendicular to the eye, I established that $2i+b=90$, therefore $b=90-2i$ which we would be substituting in our equation to find angle i. $$\frac{\sin(180-i)}{H}=\frac{\sin(a-b)}{1}$$ $$\frac{\sin(i)}{H}=\sin(a-(90-2i))=-\cos(a+2i)$$ $$\frac{\sin(i)}{H}=\sin(a)\sin(2i)-\cos(a)\cos(2i)$$ $$\sin(i)=\sin(2i)-\frac{\cos(2i)}{\sin15}$$
Here I was able to use guess and check to get the answer but I want to learn how to solve it by calculation so that I may be able to calculate other distances. I would also like to point out that $H\sin(a)=1$ and $H\cos(a)=\frac{1}{\sin15}$ based on the diagram.
$$\sin(i)=2\sin(i)\cos(i)-\frac{2\cos^2(i)-1}{\sin15}$$ $$2\sin(i)\cos(i)-\sin(i)=\frac{2\cos^2(i)-1}{\sin15}$$ $$\sin(i)=\sqrt{1-\cos^2(i)}=\frac{2\cos^2(i)-1}{\sin15(2\cos(i)-1)}$$ $$1-\cos^2(i)=\frac{(2\cos^2(i)-1)^2}{\sin^215(2\cos(i)-1)^2}$$ $$\sin^215(1-\cos^2(i))(2\cos(i)-1)^2-(2\cos^2(i)-1)^2=0$$ I'm currently stuck here as I realised I couldn't factor this equation and have no idea how to solve this from this point onward. I am still trying to understand polynomial equations and would appreciate some guidance on it. Thanks!!
Assuming that your last equation is correct and that everything is in degree, for simpler notations let $a=\sin ^2(15 {}^{\circ})$ and $c=\cos (i {}^{\circ})$.
Expanding, we have $$4( a+1) c^4-4 a c^3-(3 a+4) c^2+4 a c+(1-a)=0$$ which is quartic that, if you want nightmares, you could solve with radicals.
We have $a=\frac{2-\sqrt{3}}{4}$. Plotting the function for $-1 \leq c \leq 1$, we see four roots. Zooming more and more around the roots, we see that they are close to $-0.83$, $-0.52$, $0.68$ and $0.74$.
To polish the roots, Newton method converges in a couple of iterations and the slutions are $$c_1=-0.831525\qquad c_2=-0.523722\qquad c_3= 0.682282\qquad c_4=0.735747$$ No, do not forget that $c_k=\cos (i_k {}^{\circ})$ then $i_k {}^{\circ}$.