Math Doubts

Proof of Difference to Product identity of Sin functions

Let $x$ and $y$ be two angles of two right triangles, the sine functions with the two angles are written as $\sin{x}$ and $\sin{y}$ respectively. The transformation of difference of sine functions into product form can be proved in trigonometry by the trigonometric identities.

Subtract the sin functions for their difference

Now, subtract the sin function $\sin{y}$ from the sin function $\sin{x}$ for expressing them in subtraction form.

$\implies$ $\sin{x}-\sin{y}$

Expand each term in the expression

In fact, the trigonometric expression $\sin{x}-\sin{y}$ cannot be simplified directly but it is not impossible. Each term can be expanded by taking both angles equals to two compound angles.

Assume, $x = a+b$ and $y = a-b$. Now, replace the equivalent values of $x$ and $y$ in the trigonometric expression.

$\implies$ $\sin{x}-\sin{y}$ $\,=\,$ $\sin{(a+b)}$ $-$ $\sin{(a-b)}$

The sin functions with compound angles can be expanded by the angle sum and angle difference identities of sin functions.

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

Now, use the fundamental operations for simplifying the trigonometric expression.

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

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

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

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

Therefore, it is transformed the difference of the sin functions into product form successfully but the product form is in terms of $a$ and $b$. So, we have to convert them back into $x$ and $y$.

Get product form for difference of sin functions

In the above step, We have taken that $x = a+b$ and $y = a-b$.

The value of $x$ can be evaluated by adding both equations.

$\implies$ $x+y$ $\,=\,$ $(a+b)+(a-b)$

$\implies$ $x+y$ $\,=\,$ $a+b+a-b$

$\implies$ $x+y$ $\,=\,$ $a+a+b-b$

$\implies$ $x+y$ $\,=\,$ $\require{cancel} 2a+\cancel{b}-\cancel{b}$

$\implies$ $x+y \,=\, 2a$

$\implies$ $2a \,=\, x+y$

$\,\,\, \therefore \,\,\,\,\,\,$ $a \,=\, \dfrac{x+y}{2}$

Similarly, the value of $y$ can be calculated by subtracting the equation $y = a-b$ from the equation $x = a+b$.

$\implies$ $x-y$ $\,=\,$ $(a+b)-(a-b)$

$\implies$ $x-y$ $\,=\,$ $a+b-a+b$

$\implies$ $x-y$ $\,=\,$ $a-a+b+b$

$\implies$ $x-y$ $\,=\,$ $\require{cancel} \cancel{a}-\cancel{a}+2b$

$\implies$ $x-y \,=\, 2b$

$\implies$ $2b \,=\, x-y$

$\,\,\, \therefore \,\,\,\,\,\,$ $b \,=\, \dfrac{x-y}{2}$

We have derived that $\sin{x}-\sin{y}$ $\,=\,$ $2\cos{a}\sin{b}$ in the above step. Now, substitute the values of $a$ and $b$ in this equation.

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

Therefore, the difference of the two sin functions is transformed successfully into product form of the trigonometric functions and it is called as the difference to product identity of the sine functions.

The difference to product formula of sin functions can also be proved in terms of $C$ and $D$ by taking $C$ instead of $x$ and $D$ instead of $y$.

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

In the similar way, you can also derive the transformation of difference into product identity of sin functions in terms of any two angles.

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