The sum to product transformation identity of cos functions can be derived in trigonometry by using some trigonometric identities. Now, let’s prove it mathematically. If $x$ and $y$ represent angles of two right triangles, then the cos functions in terms of both angles are written as $\cos{x}$ and $\cos{y}$ respectively.

Now, add the cos function $\cos{x}$ to another cosine function $\cos{y}$ for expressing them in their summation form.

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

According to trigonometry, $\cos{x}+\cos{y}$ is a trigonometric expression, which cannot be simplified directly but it is not impossible. The trigonometric expression can be simplified by considering the angles in cosine functions as two compound angles.

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

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

Now, each term in the expression can be expanded by the angle sum and angle difference of cos functions.

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

Use the fundamental mathematical operations and simplify the trigonometric expression.

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

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

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

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

In this step, we have successfully transformed the sum of the cosine functions into product form but there is a problem here. Actually, the product form of the trigonometric functions should be in terms of $x$ and $y$ but they are in terms of $a$ and $b$. So, we have to eliminate them.

We can evaluate $a$ by adding the equations $x = a+b$ and $y = a-b$.

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

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

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

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

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

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

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

We can also evaluate $b$ by subtracting the equation $y = a-b$ from the equation $x = a+b$.

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

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

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

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

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

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

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

Therefore, we have successfully evaluated both $a$ and $b$. Now, substitute them in the equation $\cos{x}+\cos{y}$ $\,=\,$ $2\cos{a}\cos{b}$.

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

Therefore, the sum of the cosine functions is successfully transformed into product form of the trigonometric functions. The equation is called the sum to product identity of the cosine functions. It can also be derived in terms of $C$ and $D$ as follows by by taking $C$ instead of $x$ and $D$ instead of $y$.

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

You can also derive the transformation of sum into product identity of cosine functions in terms of any angles by following the same procedure.

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