# Proof of Difference to Product identity of Cosine functions

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

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

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

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

It is your time to learn how to prove the difference to product transformation identity for the cosine functions.

When the symbols $\alpha$ and $\beta$ denote the angles of right triangles, the cosine of angle alpha and cosine of angle beta are written as $\cos{\alpha}$ and $\cos{\beta}$ respectively in the trigonometric mathematics.

### Express the Subtraction of Cosine functions

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

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

### Expand each cosine function in the expression

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

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

As per the angle sum and angle difference trigonometric identities of cosine functions, the cosines of compound angles can be expanded mathematically in terms of trigonometric functions.

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

### Simplify the Trigonometric expression

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

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

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

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

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

Thus, we have proved the transformation of difference of the cos functions into product form successfully. Actually, the product form is derived in terms of $a$ and $b$. Therefore, we must have to express them in terms of $\alpha$ and $\beta$.

We considered that $\alpha = a+b$ and $\beta = a-b$. So, let’s evaluate both $a$ and $b$ in terms of $\alpha$ and $\beta$ by some mathematical operations.

Add the both algebraic equations for calculating the $a$ in terms of $\alpha$ and $\beta$.

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

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

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

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

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

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

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

In the same way, subtract the equation $\beta = a-b$ from the equation $\alpha = a+b$ for calculating the $b$ in terms of $\alpha$ and $\beta$.

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

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

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

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

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

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

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

We have successfully evaluated $a$ and $b$. Now, substitute them in the trigonometric equation $\cos{\alpha}-\cos{\beta}$ $\,=\,$ $-2\sin{a}\sin{b}$ for proving the difference to product identity of cosine functions.

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

Therefor, it is proved that the difference of the cosine functions is successfully converted into product form of the trigonometric functions and This trigonometric equation is called as the difference to product identity of cosine functions.

In this way, we can prove the difference to product transformation trigonometric identity of cosine functions in terms of $C$ and $D$ and also in terms of $x$ and $y$.

Latest Math Problems
Email subscription
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more