Show that `(sinx-siny)/(cosx+cosy)=tan((x-y)/2)`

Answer

To show that

\[ \frac{\sin x - \sin y}{\cos x + \cos y} = \tan\left(\frac{x - y}{2}\right), \]

we will start with the left-hand side (LHS) and manipulate it to reach the right-hand side (RHS).

### Step 1: Use the sine difference formula

We know that

\[ \sin x - \sin y = 2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right). \]

So, we can rewrite the LHS as:

\[ \frac{\sin x - \sin y}{\cos x + \cos y} = \frac{2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)}{\cos x + \cos y}. \]

### Step 2: Use the cosine sum formula

We also know that

\[ \cos x + \cos y = 2 \cos\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right). \]

Now, we can substitute this into our expression:

\[ = \frac{2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)}{2 \cos\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)}. \]

### Step 3: Simplify the expression

The \(2 \cos\left(\frac{x + y}{2}\right)\) terms in the numerator and denominator cancel out (assuming \(\cos\left(\frac{x + y}{2}\right) \neq 0\)):

\[ = \frac{\sin\left(\frac{x - y}{2}\right)}{\cos\left(\frac{x - y}{2}\right)}. \]

### Step 4: Recognize the tangent function

The expression

\[ \frac{\sin\left(\frac{x - y}{2}\right)}{\cos\left(\frac{x - y}{2}\right)} \]

is the definition of the tangent function:

\[ = \tan\left(\frac{x - y}{2}\right). \]

### Conclusion

Thus, we have shown that

\[ \frac{\sin x - \sin y}{\cos x + \cos y} = \tan\left(\frac{x - y}{2}\right), \]

which completes the proof.

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