Given: $$ \quad \cos (A-B)=\cos A \cos B+\sin A \sin B $$ Derive the formula $$ \sin (A+B)=\sin A \cos B+\cos A \sin B $$

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1. Begin with the cosine difference formula:  
   $$
   \cos (X - Y)= \cos X \cos Y + \sin X \sin Y.
   $$

2. Replace $$ X $$ with $$ \frac{\pi}{2} - A $$ and $$ Y $$ with $$ B $$. Then, the formula becomes:  
   $$
   \cos\left(\left(\frac{\pi}{2}-A\right)-B\right)
   = \cos\left(\frac{\pi}{2}-A\right) \cos B + \sin\left(\frac{\pi}{2}-A\right) \sin B.
   $$

3. Simplify the left-hand side using the property of cosine of a complementary angle:  
   $$
   \cos\left(\frac{\pi}{2}-(A+B)\right) = \sin (A+B).
   $$
   So,  
   $$
   \sin (A+B)= \cos\left(\frac{\pi}{2}-A\right) \cos B + \sin\left(\frac{\pi}{2}-A\right) \sin B.
   $$

4. Recall the co-function identities:  
   $$
   \cos\left(\frac{\pi}{2}-A\right)=\sin A,\quad \sin\left(\frac{\pi}{2}-A\right)=\cos A.
   $$

5. Substitute these identities into the expression:  
   $$
   \sin (A+B)= \sin A \cos B + \cos A \sin B.
   $$

Thus, the formula is derived:
$$
\boxed{\sin (A+B)= \sin A \cos B + \cos A \sin B.}
$$
$$ \sin (A+B) = \sin A \cos B + \cos A \sin B $$

Given: \( \quad \cos (A-B)=\cos A \cos B+\sin A \sin B \) Derive the formula \( \sin (A+B)=\sin A \cos B+\cos A \sin B \)

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