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$\sin{(A+B)}$ $+$ $\sin{(A-B)}$ formula

Formula

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

Proof

$A$ 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 $\sin{(A+B)}$ $+$ $\sin{(A-B)}$.

Expansions of sine of sum and difference angles

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.

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

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

Add both trigonometric equations

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

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

Simplify the trigonometric equation

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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