The sum to product identity of sine functions is written popularly in trigonometry in any one of the following three forms. Now, let us learn how to prove the sum to product transformation identity of sine functions.

$(1). \,\,\,$ $\sin{\alpha}+\sin{\beta}$ $\,=\,$ $2\sin{\Bigg(\dfrac{\alpha+\beta}{2}\Bigg)}\cos{\Bigg(\dfrac{\alpha-\beta}{2}\Bigg)}$

$(2). \,\,\,$ $\sin{x}+\sin{y}$ $\,=\,$ $2\sin{\Bigg(\dfrac{x+y}{2}\Bigg)}\cos{\Bigg(\dfrac{x-y}{2}\Bigg)}$

$(3). \,\,\,$ $\sin{C}+\sin{D}$ $\,=\,$ $2\sin{\Bigg(\dfrac{C+D}{2}\Bigg)}\cos{\Bigg(\dfrac{C-D}{2}\Bigg)}$

When $\alpha$ and $\beta$ represent the angles of right triangles, the sine of angle alpha is written as $\sin{\alpha}$ and sine of angle beta is written as $\sin{\beta}$ in mathematics.

Write the both sine functions in a row by displaying a plus sign between them for expressing the addition of sine functions in mathematical form.

$\implies$ $\sin{\alpha}+\sin{\beta}$

Let’s assume that $\alpha \,=\, a+b$ and $\beta \,=\, a-b$. Now, substitute them in the above trigonometric expression.

$\implies$ $\sin{\alpha}+\sin{\beta}$ $\,=\,$ $\sin{(a+b)}$ $+$ $\sin{(a-b)}$

The sines of compound angles can be expanded by the angle sum and angle difference trigonometric identities of sine functions.

$\implies$ $\sin{\alpha}+\sin{\beta}$ $\,=\,$ $\Big(\sin{a}\cos{b}$ $+$ $\cos{a}\sin{b}\Big)$ $+$ $\Big(\sin{a}\cos{b}$ $-$ $\cos{a}\sin{b}\Big)$

The trigonometric expression in the right hand side of the equation can be simplified by the fundamental mathematical operations.

$=\,\,\,$ $\sin{a}\cos{b}$ $+$ $\cos{a}\sin{b}$ $+$ $\sin{a}\cos{b}$ $-$ $\cos{a}\sin{b}$

$=\,\,\,$ $\sin{a}\cos{b}$ $+$ $\sin{a}\cos{b}$ $+$ $\cos{a}\sin{b}$ $-$ $\cos{a}\sin{b}$

$=\,\,\,$ $2\sin{a}\cos{b}$ $+$ $\require{cancel} \cancel{\cos{a}\sin{b}}$ $-$ $\require{cancel} \cancel{\cos{a}\sin{b}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\sin{x}+\sin{y}$ $\,=\,$ $2\sin{a}\cos{b}$

Therefore, it is successfully transformed the sum of the sin functions into product form but the product form is in terms of $a$ and $b$. So, they should be expressed in terms of $\alpha$ and $\beta$.

We have taken that $\alpha = a+b$ and $\beta = a-b$. So, we have to evaluate $a$ and $b$ in terms of $\alpha$ and $\beta$. They can be calculated by the fundamental operations of the mathematics.

Firstly, add both algebraic equations to find the value of $a$.

$\implies$ $\alpha+\beta$ $\,=\,$ $(a+b)+(a-b)$

$\implies$ $\alpha+\beta$ $\,=\,$ $a+b+a-b$

$\implies$ $\alpha+\beta$ $\,=\,$ $a+a+b-b$

$\implies$ $\alpha+\beta$ $\,=\,$ $2a+\cancel{b}-\cancel{b}$

$\implies$ $\alpha+\beta \,=\, 2a$

$\implies$ $2a \,=\, \alpha+\beta$

$\,\,\, \therefore \,\,\,\,\,\,$ $a \,=\, \dfrac{\alpha+\beta}{2}$

Now, subtract the equation $\beta = a-b$ from the equation $\alpha = a+b$ for getting the value of $b$ in terms of $\alpha$ and $\beta$.

$\implies$ $\alpha-\beta$ $\,=\,$ $(a+b)-(a-b)$

$\implies$ $\alpha-\beta$ $\,=\,$ $a+b-a+b$

$\implies$ $\alpha-\beta$ $\,=\,$ $a-a+b+b$

$\implies$ $\alpha-\beta$ $\,=\,$ $\cancel{a}-\cancel{a}+2b$

$\implies$ $\alpha-\beta \,=\, 2b$

$\implies$ $2b \,=\, \alpha-\beta$

$\,\,\, \therefore \,\,\,\,\,\,$ $b \,=\, \dfrac{\alpha-\beta}{2}$

We have derived that $\sin{\alpha}+\sin{\beta}$ $\,=\,$ $2\sin{a}\cos{b}$. Now, replace the values of $a$ and $b$ in the equation.

$\,\,\, \therefore \,\,\,\,\,\,$ $\sin{\alpha}+\sin{\beta}$ $\,=\,$ $2\sin{\Big(\dfrac{\alpha+\beta}{2}\Big)}\cos{\Big(\dfrac{\alpha-\beta}{2}\Big)}$

Therefore, the sum of the sine functions is successfully transformed into product form of the trigonometric functions. This trigonometric equation is called the sum to product identity of sine functions.

In this way, we can derive the sum to product transformation trigonometric identity of sine functions in terms of $x$ and $y$ and also in terms of $C$ and $D$.

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