How solve this trigo system ?

[Dieter]

(09-18-2017 08:56 AM)ggauny@live.fr Wrote:  I would like solve my exercize with my Prime but I dont' know
how to do.

The system is :

cos x - cos y = 1/2
sin x * sin y = 3/8

I have tryied all things with SOLVE but no good it is.

As Gerson already noted, you can use trigonometric identities. For instance the most simple one: sin²x + cos²x = 1 so that sin x = √(1–cos²x).
Now use this in the second equation and get (after squaring)

(1–cos²x) * (1–cos²y) = (3/8)²

Since
cos x = cos y + 1/2

you get
[1 –  (cos y + 1/2)²] * (1 – cos²y) = 9/64

I was too lazy to solve this manually, so I just fed it to the Solver of my HP35s.
Note that you do not solve for y but for cos y here. Let's call the cosine "c":

EQN
(1–(C+0,5)^2) x (1–C^2) = 9÷64

If we assume that c is somewhere between 0 and 1, these two guesses yield C=0,411437827766.

So y = arccos c = 65,704811° and x = arccos(c+1/2) = 24,295189° (which is 90°–y).

Starting with –1 and 0 as initial guesses returns C=–0,911437827766, leading to y = 155,704811° and x = 114,295189°.

No Prime required here. ;-)
But of course you can also use the Prime's solver with the above equation.

Note 1: the exact values for c are –¼ · (1±√7).

Note 2: the solver equation can be written as C⁴ + C³ – 7/4 C² – C + 39/64 = 0.
Compare this with Gerson's quartic equation above.

Dieter