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.