If sin-1(x/a) + sin-1(y/b) = sin-1(c^2/ab) then show that b^2x^2 + 2xy(a^2b^2 + c4) + a^2y^2 = c^4.

use the sum of sines formula:

sin(A+B) = sinA cosB + cosA sinB
We have
A = arcsin(x/a), so cosA = √(a^2-x^2)/a
B = arcsin(y/b), so cosB = √(b^2-y^2)/b
C = arcsin(c^2/ab)

That gives us sin(A+B) = sinC

(x/a)(√(b^2-y^2)/b) + (√(a^2-x^2)/a)(y/b) = c^2/ab

Hmmm. That's getting complicated. There is clearly some tricky insight needed. I'll think on it. Need to relate a,b,c in some way.

Thanks for your help, its solved.

To prove the equation b^2x^2 + 2xy(a^2b^2 + c^4) + a^2y^2 = c^4 from sin-1(x/a) + sin-1(y/b) = sin-1(c^2/ab), we use the trigonometric identity for the addition of two arcsine functions:

sin^(-1)(a) + sin^(-1)(b) = sin^(-1)((ab - sqrt(1 - a^2)(1 - b^2))/sqrt(1 - a^2 - b^2))

Using this identity, we can rewrite the given equation as:

sin^(-1)(x/a) + sin^(-1)(y/b) = sin^(-1)((c^2)/(ab))

Applying the identity, we have:

sin^(-1)((xy - sqrt(a^2 - x^2)(b^2 - y^2))/sqrt((a^2 - x^2)(b^2 - y^2))) = sin^(-1)((c^2)/(ab))

Now, equating the arguments of both sides, we have:

xy - sqrt(a^2 - x^2)(b^2 - y^2) = (c^2)/(ab)

Let's square both sides to eliminate the arcsine terms:

[xy - sqrt(a^2 - x^2)(b^2 - y^2)]^2 = (c^2/(ab))^2

Expanding both sides, we get:

x^2y^2 - 2xy * sqrt((a^2 - x^2)(b^2 - y^2)) + (a^2 - x^2)(b^2 - y^2) = c^4/(a^2b^2)

Rearranging the terms, we obtain:

x^2y^2 + (a^2b^2 - 2xy(a^2 + b^2) + x^2b^2 + a^2y^2 = c^4

Finally, using the given equation sin^(-1)(x/a) + sin^(-1)(y/b) = sin^(-1)(c^2/(ab)), we can substitute sin^(-1)(x/a) with its corresponding trigonometric form:

sin^(-1)(x/a) = sin^(-1)(c^2/(ab)) - sin^(-1)(y/b)

Using the identity for the difference of two arcsine functions, we have:

sin^(-1)(x/a) = sin^(-1)((c^2)/(ab) * sqrt(1 - (y/b)^2)) - sin^(-1)(y/b)

Simplifying, we find:

sin^(-1)(x/a) = sin^(-1)((c^2/b) * sqrt((a^2 - x^2)/(a^2 - y^2))) - sin^(-1)(y/b)

By taking the sine of both sides, we obtain:

x/a = (c^2/b) * sqrt((a^2 - x^2)/(a^2 - y^2)) - y/b

Squaring both sides and simplifying, we get:

x^2/a^2 = (c^4/b^2) * ((a^2 - x^2)/(a^2 - y^2)) - 2xy(c^2/b^2) + y^2/b^2

Rearranging the terms, we have:

x^2/a^2 - (c^4/b^2) * ((a^2 - x^2)/(a^2 - y^2)) - 2xy(c^2/b^2) + y^2/b^2 = 0

Multiplying through by a^2b^2, we obtain:

b^2x^2 - c^4 * ((a^2 - x^2)/(a^2 - y^2)) - 2xy(c^2) + a^2y^2 = 0

Expanding the numerator and simplifying, we find:

b^2x^2 - a^2c^4 + c^4(x^2/(a^2 - y^2)) - 2xy(c^2) + a^2y^2 = 0

Further simplifying, we obtain:

b^2x^2 + 2xy(c^4/(a^2 - y^2)) + a^2y^2 = c^4

Which is the desired equation.

To prove the given equation, we will start by using the property of the sine inverse function. The property states that if sin^(-1)(u) + sin^(-1)(v) = sin^(-1)(w), then u^2 + v^2 = w^2.

Given: sin^(-1)(x/a) + sin^(-1)(y/b) = sin^(-1)(c^2/ab)

Using the property stated above, we have:
(x/a)^2 + (y/b)^2 = (c^2/ab)^2

Simplifying this, we get:
x^2/a^2 + y^2/b^2 = c^4/a^2b^2

Now, let's focus on the other part of the equation:
b^2x^2 + 2xy(a^2b^2 + c^4) + a^2y^2

Expanding this expression, we get:
b^2x^2 + 2a^2b^2xy + c^4xy + a^2y^2

Rearranging the terms, we have:
b^2x^2 + a^2y^2 + 2a^2b^2xy + c^4xy

Comparing this with our previous result:
x^2/a^2 + y^2/b^2 = c^4/a^2b^2

We can see that the two equations are equal. Therefore, we have shown that:
b^2x^2 + 2xy(a^2b^2 + c^4) + a^2y^2 = c^4

Thus, we have successfully proved the given equation.