Math Identities problem?

Question

I am looking at an example and Cos(90 - [A + B]) changes to Sin[A + B] and I have no idea why.

Answer 1

Cos(90 - [A + B])

Identitiy: Cos(X - Y) = CosXCosY + SinXSinY

Since Cos 90° = 0

Cos(X - (A + B)) = (0)Cos(A + B) + (1)Sin(A + B)

Cos(X - (A + B)) = Sin(A + B)

Hope this helps.
.

Answer 2

Everyone is overcomplicating this.
Think of [A+B] as being an angle.

One of the values is the sine of angle [A+B], which is (by definition) the ratio of the side opposite angle [A+B] to the hypotenuse.

If one angle of a right triangle is [A+B], then the other acute angle is (90-[A+B]). And the the cosine of angle (90 - [A+B]), is the ratio of the side ADJACENT to that angle to the hypotenuse. But the side adjacent to (90 - [A+B]) is the same side that was OPPOSITE angle [A+B]. So the two ratios we're talking about are the SAME ratio. That is:
sin[A+B] = cos(90 - [A+B])

Answer 3

well the cos of 90 is 1

so its left with cos(1-[a+b]) and so it turns into
sin(a+b)