This is some trig saying that when we rotate the point around the origin by any angle the point we end up with is the same distance from the origin as the original point.
That doesn't matter. There's nothing tricky about this one, we just have to multiply it out and see where we are.
[tex] (x \cos A + y \sin A)^2 + (x \sin A - y \cos A)^2 [/tex]
[tex] = x^2 \cos^2 A + 2 xy \cos A \sin A + y^2 \sin^2 A + x^2 sin^2A - 2 x y\sin A \cos A + y^2 \cos^2 A[/tex]
Collect terms, and let the cross products (with the 2) cancel.
[tex] = x^2 (\cos^2 A + \sin ^2 A) + y^2 (\sin^2 A + \cos^2 A)[/tex]
[tex] = x^2 + y^2[/tex]
because of course [tex] \cos^2 A + \sin ^2 A = 1[/tex]
(x cosA + y sinA)^2 + (x sinA - y cosA)^2 = x^2+ y^2
LHS
(x cosA + y sinA)² + (x sinA - y cosA)²
expanding using the formula (a+b)²=a²+2ab+b² and (a-b)²=a²-2ab+b²
(x cosA)²+2(xcosA)(ysinA) + (y sinA )² + (x sinA)² - 2(x sinA)(y cosA) +(y cosA)²
x² cos²A + 2(xcosA)(ysinA) + (y² sin²A ) + (x² sin²A) - 2(x sinA)(y cosA) +(y² cos²A)
x²(cos²A+sin²A) +y²(sin²A+cos²A) +2(xcosA)(ysinA)- 2(x sinA)(y cosA)
We apply rule here sin²A+cos²A=1 and
cancelling 2(xcosA)(ysinA)- 2(x sinA)(y cosA)
we get x²+y²
RHS : x²+y²
LHS=RHS
HENCE PROVED