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Hi,

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.

May be pepol know how to do this and can help me.

thanks.
From the first equation,

cos(y)=cos(x)-1/2

(sin(y))^2=1-(cos(x))^2+cos(x)-1/4

After squaring the second equation and replacing, we get

(cos(x))^4 - (cos(x))^3 - 7/4*(cos(x))^2 + cos(x) + 39/64 = 0

which can be solved by PROOT on the 50g:

[ 1 -1 '-7/4' 1 '39/64' ]

PROOT ->

cos(x) = [ -.411437827766 .911437827766 -1.04903810568 1.54903810568 ]

The first two roots appear to be

'(1-√7)/4' and '(1+√7)/4',

respectively.

Edited to fix the sign of one of the exact solutions.
Thank Gerson,

So it is no one command which give the answer.

I lack too much mathematics to find like this !

You say first 2 roots, so there are others ?

The teacher say there is 4 solutions because I learn by correspondancy
I wait the answer of him. I shall give you him solutions.
(09-18-2017 10:45 AM)ggauny@live.fr Wrote: [ -> ]Thank Gerson,

So it is no one command which give the answer.

I lack too much mathematics to find like this !

You're welcome!

I think both the Prime and 50g can solve this, but that was what I could do without reading the manual. Other trigonometric identities might give better solutions.

Gerson.
(09-18-2017 10:45 AM)ggauny@live.fr Wrote: [ -> ]Thank Gerson,

You say first 2 roots, so there are others ?

The vector returned by PROOT contains four roots.

The other two appear to be

'(1-3*√3)/4' and '(1+3*√3)/4'.

You might want to solve the quartic equation symbolically to check the results. I fear I cannot do that by hand :-)

Notice those are solutions for cos(x). The solutions for x are obviously the arccosine of those values.

W|A lists them all:

http://m.wolframalpha.com/input/?i=solve...2F64+%3D+0
(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
(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.

It you want to use a solver there is another, shorter solution.

This is a nonlinear equation system with two unknowns. One way to solve such a system is the substitution method which has already been discussed. Another approach is solving both equations for the same variable, i.e. for x or y, and set equation1 = equation2:

y = arccos(cos x – 0,5)
y = arcsin(0,375 / sin x)

=>  arccos(cos x – 0,5) = arcsin(0,375 / sin x)

Now any decent solver should be able to solve this (x=24,2952° => y=65,7048°).
Which does not mean that any solver will return all possible solutions. ;-)

AFAIK on the Prime you can also find the intersection of both graphs, which should return the same result.

Dieter
(09-18-2017 08:56 AM)ggauny@live.fr Wrote: [ -> ]cos x - cos y = 1/2
sin x * sin y = 3/8
Alternatively, from a different angle :-), set x = p+q and y = p-q so that we get

-2sin(p)sin(q) = 1/2
(sin(p)cos(q) + cos(p)sin(q)) * (sin(p)cos(q) - cos(p)sin(q)) = 3/8

and eliminating cos^2 for sin^2 after simplifying the latter equation we get

-sin(p)sin(q) = 1/4
sin(p)^2-sin(q)^2 = 3/8

So you can solve this exactly without the ugliness of dealing with a quartic....
(09-18-2017 08:56 AM)ggauny@live.fr Wrote: [ -> ]cos x - cos y = 1/2
sin x * sin y = 3/8

Finally, you may try wolframalpha.com to solve this. Simply enter
solve (cos x - cos y = 1/2) AND (sin x * sin y = 3/8) for x,y

The returned results include some nice (x,y) plots that show multiple sets of two ring-shaped graphs with four intersections each. Very illustrative, I think.

Dieter
Hi,

Thanks at all for your explainations. In attachement is the solution

Have a good day.
(09-22-2017 07:27 AM)ggauny@live.fr Wrote: [ -> ]Thanks at all for your explainations. In attachement is the solution

“Impossible to go further”. Not impossible at all, perhaps only more difficult to do it by hand that way.

Another way, starting from the second try, not very different from what has been suggested so far.
(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:

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

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

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°
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