Math Doubts

Proof of Difference to Product identity of Cos functions

Assume, $x$ and $y$ represent two angles of two right triangles, then the cosine functions with those angles are written as $\cos{x}$ and $\cos{y}$ respectively. Now, let us derive the transformation of difference of cos functions into product form mathematically in trigonometry by using the trigonometric identities.

Subtract the cos functions for their difference

Subtract the cosine function $\cos{y}$ from the cos function $\cos{x}$ for expressing the relation between them in subtraction form.

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

Expand each term in the expression

Actually, it is not possible to simplify the trigonometric expression $\cos{x}-\cos{y}$ directly but it is not impossible. There is a possibility to simplify it if we take the angles in the cos functions as the compound angles.

Let us take $x = a+b$ and $y = a-b$. Now, replace the $x$ and $y$ by their equivalent values in the cos functions.

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

According to the angle sum and angle difference identities of cos functions, the cosine functions that contain compound angles can be expanded mathematically.

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

The trigonometric expression can be simplified further by using the fundamental mathematical operations.

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

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

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

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

Therefore, the trigonometric expression in terms of cos functions is successfully transformed into product form but the product form is obtained in terms of $a$ and $b$. Therefore, we have to transform them back into $x$ and $y$.

Get product form for difference of cos functions

We have taken above that $x = a+b$ and $y = a-b$. The value of $x$ can be calculated by adding the both equations.

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

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

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

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

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

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

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

In the same way, the value of $y$ can be evaluated by subtracting the equation $y = a-b$ from the equation $x = a+b$.

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

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

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

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

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

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

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

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

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

Therefore, the difference of the two cosine functions is successfully transformed into product form of the trigonometric functions. The difference to product form equation is called as the difference to product identity of the cosine functions.

The formula for the difference to product transformation of cosine functions can be proved in terms of $C$ and $D$ by taking $C$ instead of $x$ and $D$ instead of $y$.

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

In this way, you can prove the transformation identity of difference into product form for the cosine functions in terms of any two angles.

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