Question 4 this is addma under trigonometry compound and multiple angles

We use the compound angle formulas:
$\sin(A+B) = \sin A \cos B + \cos A \sin B$
$\sin(A-B) = \sin A \cos B - \cos A \sin B$
Subtracting the second from the first:

$$\sin(A+B) - \sin(A-B) = (\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B)$$ $$= \sin A \cos B + \cos A \sin B - \sin A \cos B + \cos A \sin B$$ $$= \boxed{2 \cos A \sin B}$$

We use the compound angle formulas:
$\cos(A+B) = \cos A \cos B - \sin A \sin B$
$\cos(A-B) = \cos A \cos B + \sin A \sin B$
Subtracting the second from the first:

$$\cos(A+B) - \cos(A-B) = (\cos A \cos B - \sin A \sin B) - (\cos A \cos B + \sin A \sin B)$$ $$= \cos A \cos B - \sin A \sin B - \cos A \cos B - \sin A \sin B$$ $$= \boxed{-2 \sin A \sin B}$$

From Step 1, we have $\sin(A+B) - \sin(A-B) = 2 \cos A \sin B$.
Substitute $X = A+B$, $Y = A-B$, $A = \frac{X+Y}{2}$, and $B = \frac{X-Y}{2}$ into this equation:

$$\sin X - \sin Y = 2 \cos \left(\frac{X+Y}{2}\right) \sin \left(\frac{X-Y}{2}\right)$$

This matches the required formula.

$$\sin X - \sin Y = \boxed{2 \sin \frac{1}{2}(X-Y) \cos \frac{1}{2}(X+Y)}$$

First, we need to simplify $\cos(A+B) + \cos(A-B)$:

$$\cos(A+B) + \cos(A-B) = (\cos A \cos B - \sin A \sin B) + (\cos A \cos B + \sin A \sin B)$$ $$= \cos A \cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B$$ $$= 2 \cos A \cos B$$

Now, substitute $X = A+B$, $Y = A-B$, $A = \frac{X+Y}{2}$, and $B = \frac{X-Y}{2}$ into this result:

$$\cos X + \cos Y = 2 \cos \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)$$

This matches the required formula.

$$\cos X + \cos Y = \boxed{2 \cos \frac{1}{2}(X+Y) \cos \frac{1}{2}(X-Y)}$$