solve the simultaneous equations sin(x)+sin(y)=sin(c) and cos(x)+cos(y)=cos(c)

Question

trigonometry equations

Answer 1

sin(A+B)+sin(A-B)=sinAcosB+cosAsinB+sinAcosB-cosAsinB=2sinAcosB.

cos(A+B)+cos(A-B)=cosAcosB-sinAsinB+cosAcosB+sinAsinB=2cosAcosB.

If A+B=x and A-B=y, then 2A=x+y, so A=½(x+y) and B=½(x-y), therefore:

(a) sin(x)+sin(y)=2sin(½(x+y))cos(½(x-y))=sin(c) and:

(b) cos(x)+cos(y)=2cos(½(x+y))cos(½(x-y))=cos(c).

(a)²+(b)²=4sin²(½(x+y))cos²(½(x-y))+4cos²(½(x+y))cos²(½(x-y))=

4cos²(½(x-y))(sin²(½(x+y))+cos²(½(x+y))=4cos²(½(x-y))=sin²(c)+cos²(c)=1,

cos(½(x-y))=±½, ½(x-y)=π/3 or 2π/3.

Also, (a)/(b)=tan(½(x+y))=tan(c), ½(x+y)=c.

Now we have a system of two equations and two unknowns:

½(x-y)=π/3 or 2π/3,

½(x+y)=c.

So x=c+π/3 or c+2π/3, and y=c-π/3 or c-2π/3.

When x=c+π/3 and y=c-π/3, sin(½(x+y))=sin(c) and cos(½(x-y))=½, so 2sin(½(x+y))cos(½(x-y))=sin(c)✔️

(b) is also true for these values of x and y.

However, when x=c+2π/3 and y=c-2π/3, sin(½(x+y))=sin(c) and cos(½(x-y))=-½, so 2sin(½(x+y))cos(½(x-y))=2sin(c)(-½)=-sin(c)❌

SOLUTION: x=c+π/3, y=c-π/3. Because of x-y symmetry x=c-π/3, y=c+π/3 is also a solution. Graphically, there are two intersection points for every instance of the pair of curves (a pair of intersecting closed loops). The x and y coordinates each differ by ⅔π. The graph consists of an endless array of pairs of intersecting closed loops.

Note that there are other solutions to the system of equations because of the cyclic nature of sine and cosine. Adding 2πn, where n is an integer to each of the values of x and y generates a whole sequence of solutions, hence the graphical array.

For example, x=c+π/3+2π, y=c-π/3+6π would also be a solution.