cos(A + B) and cos(A - B)

The cosine formulas pair cosine with cosine and sine with sine, and the sign flips:

$$\cos\left(A+B\right) = \cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)$$

$$\cos\left(A-B\right) = \cos\left(A\right)\cos\left(B\right)+\sin\left(A\right)\sin\left(B\right)$$

A sum of angles gives a minus, a difference gives a plus. A quick check: with B = 0 both reduce to cos A, and with A = B the second gives cos 0 = cos²A + sin²A = 1.

For 15° as 45° - 30°: $\cos\left(45^{\circ}-30^{\circ}\right) = \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2}$, so $\cos\left(15^{\circ}\right) = \frac{\sqrt{6}+\sqrt{2}}{4}$. That is the same number as sin 75°, as the cofunction identity says it must be: 15° and 75° add to 90°.