Final answer:
By squaring x, multiplying by y, expanding, and simplifying using trigonometric identities, we are able to prove that if x = bcos A - asin A and y = acos A + bsin A, then x²y equates to a²b².
Step-by-step explanation:
We are given x = bcos A - asin A and y = acos A + bsin A, and we need to prove that x²y = a²b². To do this, we will use trigonometric identities and algebraic manipulation.
First, let's square x:
x² = (bcos A - asin A)²
= b²cos² A - 2abcos Asin A + a²sin² A
Now, we multiply this by y:
x²y = (b²cos² A - 2abcos Asin A + a²sin² A)(acos A + bsin A)
Expand it out:
= a³b²cos³ A + a²b³cos² Asin A - 2a²b²cos Asin A - 2ab³cos² Asin A + ab³sin³ A + a³b²sin² A cos A
Notice how the terms a²b³cos² Asin A and 2ab³cos² Asin A, and also 2a²b²cos Asin A and a³b²sin² A cos A cancel each other out. This leaves us with:
= a³b²cos³ A + ab³sin³ A
We can now factor out a³b²:
= a³b²(cos³ A + sin³ A)
Using the Pythagorean identity cos² A + sin² A = 1, we can say that cos³ A + sin³ A = 1, which leaves us with:
x²y = a³b²
x²y = a²b².