If sin^-1(x/a) + sin^-1(y/b)= sin^-1(c^2/ab) then prove that b^2x^2 + a^2y^2+2xy√(c^4-a^2b^2)=c^4

Answer 1

Let
A = sin^-1(x/a)
B = sin^-1(y/b)
sin(A+B) = sinAcosB + cosAsinB
So, now we have
x/a √(b^2-y^2)/b + √(a^2-x^2)/a y/b = c^2/ab
or, placing all over a common denominator
x√(b^2-y^2) + y√(a^2-x^2) = c^2
which is just what they have if you massage it a bit

Answer 2

sin^-1(x/a) + sin^-1(y/b)= sin^-1(c^2/ab)
remember that something like sin^-1(x/a) is an angle Ø, so that sinØ = x/a

so why not take sin of both sides

SIN(sin^-1(x/a) + sin^-1(y/b) )= SIN(sin^-1(c^2/ab) )
sin(sin^-1(x/a)cos(sin^-1 (y/b) + cos(sin^-1 (x/a)sin(sin^-1 (y/b)) = c^2/ab
x/a √(b^2 - y^2)/b + √(a^2 - x^2)/a (y/b) = c^2/ab
multiply each term by ab
x√(b^2 - y^2) + y√(a^2 - x^2) = c^2
square both sides
x^2 (b^2 - y^2) + 2xy√(b^2 - y^2)(a^2 - x^2)) + y^2(a^2 - x^2) = c^4
x^2 b^2 - x^2 y^2 + 2xy√(a^2b^2 - b^2x^2 - a^2y^2 + x^2y^2) + a^2y^2 - x^2y^2= c^4

argghhhh!!!!
looking at what I am to prove, I see a c^4 in the square root.
Where did that c come from???
check my arithmetic, I should have done it on paper.