Prove algebraically the the following equation is an identity: (2 cos x + 3 sin x) ^2 + ( 3 cos x - 2 sin x)^2 =13
sorry i just updated the question b/c i was trying re edit this question with a plus sign in one of the equation so people don't get confused. but i couldn't find the edit button so just copied and pasted it again but updated.

(2cos(x) + 3sin(x))² + (3cos(x) - 2sin(x))² = 13
(2cos(x) + 3sin(x))(2cos(x) + 3sin(x)) + (3cos(x) - 2sin(x))(3cos(x) - 2sin(x)) = 13
(2cos(x)(2cos(x) + 3sin(x)) + 3sin(x)(2cos(x) + 3sin(x))) + (3sin(x)(3cos(x) - 2sin(x)) - 2sin(x)(3cos(x) - 2sin(x))) = 13
(2cos(x)(2cos(x)) + 2cos(x)(3sin(x)) + 3sin(x)(2cos(x)) + 3sin(x)(3sin(x))) + (3cos(x)(3cos(x)) - 3cos(x)(2sin(x)) - 2sin(x)(3cos(x)) + 2sin(x)(2sin(x))) = 13
(4cos²(x) + 3sin(2x) + 3sin(2x) + 9sin²(x)) + (9cos²(x) - 3sin(2x) - 3sin(2x) + 4sin²(x)) = 13
  (4cos²(x) + 6sin(2x) + 9sin²(x)) + (9cos²(x) - 6sin(2x) + 4sin²(x)) = 13
(4cos²(x) + 9cos²(x)) + (6sin(2x) - 6sin(2x)) + (9sin²(x) + 4sin²(x)) = 13
                                                                     13cos²(x) + 13sin²(x) = 13
                                       13(¹/₂(1 + cos(2x))) + 13(¹/₂(1 - cos(2x))) = 13
               13(¹/₂(1)) + 13(¹/₂(cos(2x))) + 13(¹/₂(1)) - 13(¹/₂(cos(2x))) = 13
                             13(¹/₂) + 13(¹/₂cos(2x)) + 13(¹/₂) - 13(¹/₂cos(2x)) = 13
                                               6¹/₂ + 6¹/₂cos(2x) + 6¹/₂ - 6¹/₂cos(2x) = 13
                                                6¹/₂ + 6¹/₂ + 6¹/₂cos(2x) - 6¹/₂cos(2x) = 13
                                                                                              13 + 0 = 13
                                                                                                    13 = 13
The following equation is an identity.

1.3 and 1 1/3 are both closest to 1. hope this ; )