How solve this trigo system ?

[Albert Chan]

(09-18-2017 08:56 AM)ggauny@live.fr Wrote:  cos x - cos y = 1/2
sin x * sin y = 3/8

Nice problem !
Square both side, both equations:

\( \cos^2 x - 2 \cos x \cos y + \cos^2 y = 1/4 \)
\((1-\cos^2 x)(1-\cos^2 y) = 1 - (\cos^2 x + \cos^2 y) + (\cos^2 x \cos^2 y) = 9/64 \)

Add both equations, and simplify:

\((\cos x \cos y)^2 - 2( \cos x \cos y) + 39/64 = 0 \quad\; → \cos x \cos y = 3/8\)

Using angle sum formula, we have:

\(\cos(x + y) = \cos x \cos y - \sin x \sin y = 0 \quad\quad → x + y = ±\,90° \)

This meant we can replace sin(y) as ±cos(x)

2 sin(x) cos(x) = sin(2x) = ±3/4

The process of squaring may introduce some bad solutions.
With those eliminated, we have x ≈ ±24.295°, ±114.295°
These solutions corresponded to y ≈ ±65.705°, ±155.705°