Prove sinA + sinB = 2sin((A+B)/2)cos((A-B)/2) using sum-to-product formula

Question

Prove that $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ .

(NCERT Class 11, Chapter 3 — Trigonometric Functions)

Solution — Step by Step

We’ll prove this by expanding the right-hand side using the product formula:

\sin C \cos D = \frac{1}{2}[\sin(C+D) + \sin(C-D)]

Let $C = \frac{A+B}{2}$ and $D = \frac{A-B}{2}$ .

C + D = \frac{A+B}{2} + \frac{A-B}{2} = \frac{2A}{2} = A

C - D = \frac{A+B}{2} - \frac{A-B}{2} = \frac{2B}{2} = B

2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) = 2 \cdot \frac{1}{2}[\sin(A) + \sin(B)]

= \sin A + \sin B = \text{LHS}

\boxed{\text{Hence proved: } \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}

Why This Works

The sum-to-product formulas are essentially the product-to-sum formulas read backwards. The product formula $\sin C \cos D = \frac{1}{2}[\sin(C+D) + \sin(C-D)]$ is derived from the sine addition formulas. By substituting $C = (A+B)/2$ and $D = (A-B)/2$ , we recover the original variables $A$ and $B$ .

The transformation converts a sum of sines (hard to simplify further) into a product (easier to factorise and solve equations). This is why sum-to-product formulas are so useful in solving trigonometric equations — they turn addition into multiplication.

Alternative Method — Starting from the LHS

Let $A = C + D$ and $B = C - D$ , so $C = (A+B)/2$ and $D = (A-B)/2$ .

Using the sine addition formulas:

\sin A = \sin(C+D) = \sin C \cos D + \cos C \sin D

\sin B = \sin(C-D) = \sin C \cos D - \cos C \sin D

Adding these:

\sin A + \sin B = 2\sin C \cos D = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)

For JEE, memorise all four sum-to-product formulas:

$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
$\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$
$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$

The sign in the last formula catches many students — $\cos A - \cos B$ has a negative sign out front.

Common Mistake

When using these formulas, students often get $\frac{A+B}{2}$ and $\frac{A-B}{2}$ mixed up inside $\sin$ and $\cos$ . For $\sin A + \sin B$ , the sine takes $(A+B)/2$ and the cosine takes $(A-B)/2$ . For $\sin A - \sin B$ , it’s the opposite: cosine takes $(A+B)/2$ and sine takes $(A-B)/2$ . Getting these swapped gives a wrong identity.

Question

Solution — Step by Step

Why This Works

Alternative Method — Starting from the LHS

Common Mistake

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