Math Doubts

Proof of Sum to Product identity of Sin functions

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

Add the sin functions for their sum

Now, add the sin functions $\sin{x}$ and $\sin{y}$ for expressing them in summation form.

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

Expand each term in the expression

Actually, it is not possible to simplify the trigonometric expression $\sin{x}+\sin{y}$ directly but it is not impossible. If we make each term expandable, then it is possible to convert them into product form.

Take, $x = a+b$ and $y = a-b$. Now, substitute the equivalent values in the trigonometric expression.

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

According to the angle sum and angle difference of sin functions, the functions $\sin{(a+b)}$ and $\sin{(a-b)}$ can be expanded.

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

Now, simplify the trigonometric expression 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 $x$ and $y$.

Get product form for sum of sin functions

We have taken that $x = a+b$ and $y = a-b$ in the second step of the derivation.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

In the previous step, we have derived that $\sin{x}+\sin{y}$ $\,=\,$ $2\sin{a}\cos{b}$. Now, replace the values of $a$ and $b$ in the equation.

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

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

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

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

Similarly, you can also derive the transformation of sum into product identity of sin functions in terms of any angles.

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