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

$A$ and $B$ are two angles. Sine of sum of angles is $\sin{(A+B)}$ and sine of difference of angles is $\sin{(A-B)}$. The sum of them is $\sin{(A+B)}$ $+$ $\sin{(A-B)}$.

According to angle sum and angle difference identities, $\sin{(A+B)}$ and $\sin{(A-B)}$ can be expanded in terms of sin and cos of angles.

$(1) \,\,\,\,\,\,$ $\sin{(A+B)}$ $\,=\,$ $\sin{A}\cos{B}$ $+$ $\cos{A}\sin{B}$

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

Now, add both trigonometric equations for obtaining sum of them.

$\implies$ $\sin{(A+B)}$ $+$ $\sin{(A-B)}$ $\,=\,$ $\sin{A}\cos{B}$ $+$ $\cos{A}\sin{B}$ $+$ $\sin{A}\cos{B}$ $-$ $\cos{A}\sin{B}$

Finally, simplify the trigonometric equation and obtain the sum of sine of sum of angles and sine of difference of angles in product form of the trigonometric functions.

$\implies$ $\sin{(A+B)}$ $+$ $\sin{(A-B)}$ $\,=\,$ $\sin{A}\cos{B}$ $+$ $\sin{A}\cos{B}$ $+$ $\cos{A}\sin{B}$ $-$ $\cos{A}\sin{B}$

$\implies$ $\sin{(A+B)}$ $+$ $\sin{(A-B)}$ $\,=\,$ $2\sin{A}\cos{B}$ $+$ $\require{cancel} \cancel{\cos{A}\sin{B}}$ $-$ $\require{cancel} \cancel{\cos{A}\sin{B}}$

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

Therefore, it is proved that the sum of sin functions is written as the product of the trigonometric functions sine and cosine. Hence, it is called as sum to product identity in trigonometry.

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