Question

mathematica解方程

Solve[{x == R Cos[alpha] - R Cos[alpha + beta], y == R Sin[alpha] - R Sin[alpha + beta]}, {alpha, beta}]
得不到任何结果。这是为什么呢？
怎么化简

Answer 1

我怎么有结果啊——
{{beta -> -ArcCos[(2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> -ArcCos[(R x^3 + R x y^2 - Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}, {beta -> -ArcCos[(
2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> ArcCos[(R x^3 + R x y^2 - Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}, {beta -> -ArcCos[(
2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> -ArcCos[(R x^3 + R x y^2 + Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}, {beta -> -ArcCos[(
2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> ArcCos[(R x^3 + R x y^2 + Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}, {beta ->
ArcCos[(2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> -ArcCos[(R x^3 + R x y^2 - Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}, {beta ->
ArcCos[(2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> ArcCos[(R x^3 + R x y^2 - Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}, {beta ->
ArcCos[(2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> -ArcCos[(R x^3 + R x y^2 + Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}, {beta ->
ArcCos[(2 R^2 - x^2 - y^2)/(2 R^2)],
alpha -> ArcCos[(R x^3 + R x y^2 + Sqrt[
4 R^4 x^2 y^2 - R^2 x^4 y^2 + 4 R^4 y^4 - 2 R^2 x^2 y^4 -
R^2 y^6])/(2 (R^2 x^2 + R^2 y^2))]}}

Answer 2

含有三角函数的有多解问题，应该用Reduce命令来解，你若用Solve，那么只会出现0附近的一个周期内的那个解。
a = Reduce[{x == R Cos[alpha] - R Cos[alpha + beta],
y == R Sin[alpha] - R Sin[alpha + beta]}, {alpha, beta}]

Simplify@a