Establish the identity

(2 cos delta - 3 sin delta)^2 + (3 cos delta + 2 sin delta)^2 = 13

Now I think I know how to solve for this (I'll provide my work underthis).. but Mathlab is asking me first to fill in this box on the left side??

"Rewrite the left side by removing the parentheses.

___ + 9 cos^2 delta + 12cos delta sin delta + 4sin^2 delta

I have no idea what just happened or how this broke up??

....anywho, the example later said to square everything, so I do.. This is my work.

4 cos^2 delta + 9sin^2 delta + 9 cos^2 delta + 4 sin^2 delta =

13 cos^2 delta + 13 sin^2 delta =
13 ( cos^2 delta + sin^2 delta) = ................Pythagorean: [cos^2 + sin^2 = 1].. so
13 (1) = 13
13 = 13

Don't forget the sinθcosθ term when you expand...

So,

LHS
(2cosθ - 3sinθ)² + (3cosθ + 2sinθ)²
4cos²θ - 12sinθcosθ + 9sin²θ + 9cos²θ + 12sinθcosθ + 4sin²θ
4(cos²θ + sin²θ) + 9(sin²θ + cos²θ) + (-12sinθcosθ + 12sinθcosθ)
4(1) + 9(1) + 0
13
RHS

The sinθcosθ's ended up canceling out, but don't completely ignore them as they won't always cancel out.

EDIT: Sounds like expanding may be the problem...remember:

(a + b)² = (a + b)(a + b) = a² + ab + ab + b² = a² + 2ab + b²
(a - b)² = (a - b)(a - b) = a² - ab - ab + b² = a² - 2ab + b²