Mathematica Asked by user64494 on December 24, 2020
The system is as follows
sys={a*Tan[x] == b*Tan[y], a*Sin[x]^2 + b*Cos[x]^2 == m, b*Sin[y]^2 + a*Cos[y]^2 == n}
My approachs are
Eliminate[{a*Tan[x] == b*Tan[y], a*Sin[x]^2 + b*Cos[x]^2 == m, b*Sin[y]^2 + a*Cos[y]^2 == n}, {x, y}]
Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information.
m (-a b + a n + b n) == a b n
and (The conditions a != b, a != m, b != n
are added to exclude degenerate cases.)
Reduce[{a^2*Tan[x]^2 == b^2*Tan[y]^2, a*Sin[x]^2 + b*Cos[x]^2 == m,
b*Sin[y]^2 + a*Cos[y]^2 == n, a != b, a != m, b != n}, {x, y}, Reals] // FullSimplify
A huge expression
and
Solve[{a*Tan[x] == b*Tan[y], a*Sin[x]^2 + b*Cos[x]^2 == m,
b*Sin[y]^2 + a*Cos[y]^2 == n, a != b, a != m, b != n}, {x, y}]
{}
The answer by hand under the conditions a != b, a != m, b != n
is $$ a^2(m-b)(b-n)=b^2(n-a)(a-m).$$
Weierstrass substitution
makes Eliminate
working:
Solutions of sys
are 2Pi-periodic
sys={a*Tan[x] == b*Tan[y], a*Sin[x]^2 + b*Cos[x]^2 == m, b*Sin[y]^2 +a*Cos[y]^2 == n}
Weierstrass-substitution constraints the solution {x,y}
to -Pi<x,y<Pi
sysu = TrigExpand[sys /. {x -> 2 ArcTan[ux], y -> 2 ArcTan[uy]}] // Simplify
cond=Eliminate[sysu, {ux, uy}] // FullSimplify
(*a b (m + n) == (a + b) m n*)
addendum (3.10.2020)
The magic answer by hand $$a^2 (m - b) (b - n) -b^2 (n - a) (a - m)$$ can easily be simplified to
0==Factor[a^2 (m [Minus] b) (b [Minus] n) -b^2 (n [Minus] a) (a [Minus] m)] // FullSimplify
(*(a - b) (-a m n - b m n + a b (m + n))==0*)
The second part is equivalent to the condition found by Weierstrass-substitution!
Answered by Ulrich Neumann on December 24, 2020
Only provide another way similar to Eliminate
. Not so beautiful.
Here we add extra condition.
a ∈ Reals, b ∈ Reals, x ∈ Reals, y ∈ Reals, m ∈ Reals, n ∈ Reals, a != b, a != m, b != n, m != b, n != a
Clear["`*"];
sys = {a*Tan[x] == b*Tan[y], a*Sin[x]^2 + b*Cos[x]^2 == m,
b*Sin[y]^2 + a*Cos[y]^2 == n, a ∈ Reals,
b ∈ Reals, x ∈ Reals, y ∈ Reals,
m ∈ Reals, n ∈ Reals, a != b, a != m, b != n,
m != b, n != a};
reg = ImplicitRegion[sys , {a, b, x, y, m, n}];
Resolve[Exists[{x, y}, Element[{a, b, x, y, m, n}, reg]],
Reals] // Simplify
Answered by cvgmt on December 24, 2020
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