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.