Ratio of sines given, find ratio of cosines?

Question

Is it possible to find $\displaystyle{\cos x \over \cos y}$ if $\displaystyle{\sin x\over \sin y}$ is given? If so, how would one approach this problem? Thank you in advance!

Answer 1

Here's a picture of the reverse problem.

And here's an overly-long discussion of that picture:

Given a point $P$ at some distance (here, greater than $1$) from center $O$ of the unit circle, we connect $P$ to some point $A$ on the circle. Drop a perpendicular from $O$ to $B$ on $\overline{AP}$, and let point $X$ on that perpendicular be such that $\overleftrightarrow{AX}\parallel\overleftrightarrow{OP}$.

Defining $\alpha := \angle AOB$ and $\beta = \angle BOP = \angle BXA$, we have $$\cos\alpha = \frac{|\overline{OB}|}{|\overline{OA}|} \qquad \cos\beta = \frac{|\overline{OB}|}{|\overline{OP}|}\qquad\to\qquad |\overline{OP}| = \frac{\cos\alpha}{\cos\beta}$$ Thus, $|\overline{OP}|$ "gives" a ratio of cosines. However, we also have $$\sin\alpha = \frac{|\overline{AB}|}{|\overline{OA}|} \qquad \sin\beta = \frac{|\overline{AB}|}{|\overline{AX}|} \qquad\to\qquad |\overline{AX}| = \frac{\sin\alpha}{\sin\beta}$$ Thus, $|\overline{AX}|$ is the corresponding ratio of sines; or, rather, a corresponding ratio of sines. If we choose to connect $P$ to an alternative point $A$ ---say, $A^\prime$--- an identical construction can give an entirely different ratio of sines $|\overline{A^\prime X^\prime}| = \frac{\sin\alpha^\prime}{\sin\beta^\prime} \neq \frac{\sin\alpha}{\sin\beta}$ for the same ratio of cosines $\frac{\cos\alpha^\prime}{\cos\beta^\prime} = |\overline{OP}| = \frac{\cos\alpha}{\cos\beta}$.

Answer 2

No, the ratio of sines does not uniquely determine the ratio of cosines. Let the ratio of sinse be $r$, and consider the square of the cosine ratio: $$ \frac{\cos x}{\cos y} = \frac{1-\sin^2x}{1-r^2\sin^2 x} $$ For a fixed $r$, $\frac{1-\sin^2x}{1-r^2\sin^2 y}$ will vary from zero if $\sin^2 x = 0$ to $\frac{1}{1-r^2}$ if $\sin^2 x = 1$.