((sin(a)cos(b)-cos(a)sin(b))^2)+((cos(a)cos(b)+sin(a)sin(b)^2)

Question

I need some help with the steps to solving this, I know that I need to foil out the problem but I need to get a numerical value for the expression. 

Answer 1

Let's tackle each set of parentheses separately first.

(sin a • cos b - cos a • sin b)² = (sin a • cos b - cos a • sin b)(sin a • cos b - cos a • sin b)

                                             = sin²a • cos²b - sin a cos a • sin b cos b - sin a cos a • sin b cos b + cos²a • sin²b

                                             = sin² a • cos² b - 2 (sin a cos a • sin b cos b) + cos²a • sin²b

 

Other parentheses: 

(cos a cos b + sin a sin b)² = (cos a cos b + sin a sin b)(cos a cos b + sin a sin b)

                                         = cos²a • cos² b + sin a cos a • sin b cos b + sin a cos a • sin b cos b + sin²a • sin²b

                                         = cos²a • cos²b + 2 (sin a cos a • sin b cos b) + sin² a • sin² b

 

Comparing both sets, we see the - 2 (sin a cos a • sin b cos b) will cancel out the + 2 (sin a cos a • sin b cos b) when added. So, we can omit that. We get:

 

sin²a • cos² b + cos² a • sin² b + cos² a • cos² b + sin² a • sin² b

 

Let's factor out like terms. We get:

 

sin² a (sin² b + cos² b) + cos² a (sin² b + cos² b) = sin² a + cos² a = 1

 

(using the trigonometric identity that sin² b + cos² b = 1)

 

By the way (sin a • cos b - cos a • sin b)² = sin (a-b)² and (cos a cos b + sin a sin b)² = cos (a-b)²

sin (a-b)² + cos (a-b)² = 1

Answer 2

Use these trigonometric identities:

 

sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

 

Then you are left with

 

sin²(a-b) + cos²(a-b) = 1 since sin²x + cos²x= 1