How can you prove that cos ((a+b) /2) = (sina -sinb) / (2sin ((a-b/2)?

 How can you prove that cos ((a+b) /2) = (sina -sinb) / (2sin ((a-b/2)?👇🏾👇 🏾

prove the trigonometric identity

$$\cos \left(\frac{𝑎+𝑏}{2}\right)=\frac{\sin 𝑎-\sin 𝑏}{2\sin \left(\frac{𝑎-𝑏}{2}\right)},$$

we can use some trigonometric identities and manipulations. Here's the step-by-step proof:

1. Start with the sum-to-product identities for sine:

   $$\sin 𝑎-\sin 𝑏=2\cos \left(\frac{𝑎+𝑏}{2}\right)\sin \left(\frac{𝑎-𝑏}{2}\right)\mathrm{.}$$

2. Substitute this identity into the right-hand side of the equation we want to prove:

   $$\frac{\sin 𝑎-\sin 𝑏}{2\sin \left(\frac{𝑎-𝑏}{2}\right)}\mathrm{.}$$

3. Replace $\sin 𝑎-\sin 𝑏$ using the sum-to-product identity:

   $$\sin 𝑎-\sin 𝑏=2\cos \left(\frac{𝑎+𝑏}{2}\right)\sin \left(\frac{𝑎-𝑏}{2}\right)\mathrm{.}$$

4. Now, substitute this expression into the right-hand side of our equation:

   $$\frac{2\cos \left(\frac{𝑎+𝑏}{2}\right)\sin \left(\frac{𝑎-𝑏}{2}\right)}{2\sin \left(\frac{𝑎-𝑏}{2}\right)}\mathrm{.}$$

5. Simplify the fraction:

   $$\frac{2\cos \left(\frac{𝑎+𝑏}{2}\right)\sin \left(\frac{𝑎-𝑏}{2}\right)}{2\sin \left(\frac{𝑎-𝑏}{2}\right)}=\cos \left(\frac{𝑎+𝑏}{2}\right)\mathrm{.}$$

1. Notice that the $\sin \left(\frac{𝑎-𝑏}{2}\right)$ terms cancel out:

   $$\cos \left(\frac{𝑎+𝑏}{2}\right)\mathrm{.}$$

Thus, we have shown that

$$\cos \left(\frac{𝑎+𝑏}{2}\right)=\frac{\sin 𝑎-\sin 𝑏}{2\sin \left(\frac{𝑎-𝑏}{2}\right)},$$

which completes the proof.