The difference to product identity of sine functions is written popularly in trigonometry in the following three forms.

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

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

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

Now, let us learn how to derive the difference to product transformation identity of sine functions.

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

Write the difference of the sine functions in a row by displaying a minus sign between them for expressing the subtraction of sine functions mathematically in trigonometry.

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

Assume that $\alpha \,=\, a+b$ and $\beta \,=\, a-b$. Now, replace them in the above trigonometric expression.

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

According to the angle sum and angle difference trigonometric identities of sine functions, the sines of compound angles can be expanded.

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

For getting the difference of sine functions, the trigonometric expression in the right hand side of the equation should be simplified mathematically by the fundamental operations of mathematics.

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

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

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

$\,\,\, \therefore \,\,\,\,\,\,$ $\sin{\alpha}-\sin{\beta}$ $\,=\,$ $2\cos{a}\sin{b}$

Therefore, it is proved the transformation of difference of the sin functions into product form successfully. However, the product form is derived in terms of $a$ and $b$. Hence, we should have to express them in terms of $\alpha$ and $\beta$.

You know that we have considered $\alpha = a+b$ and $\beta = a-b$. Now, let’s evaluate the $a$ and $b$ in terms of $\alpha$ and $\beta$ by the mathematical operations.

Now, add the both algebraic equations for evaluating the $a$.

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

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

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

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

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

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

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

Similarly, subtract the equation $\beta = a-b$ from the equation $\alpha = a+b$ for evaluating the $b$.

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

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

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

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

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

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

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

We have simplified the trigonometric equation above that $\sin{\alpha}-\sin{\beta}$ $\,=\,$ $2\cos{a}\sin{b}$. We can now replace the values of $a$ and $b$ in the equation.

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

Therefore, the difference of the sine functions is successfully converted into product form of the trigonometric functions. The trigonometric equation is called as the difference to product identity of sine functions.

Similarly, you can derive the difference 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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