Math Doubts

Proof of Sum to Product identity of Cosine functions

The sum to product identity of cosine functions is written in trigonometry popularly in any one of the following three forms.

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

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

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

Now, let’s learn how to derive the sum to product transformation identity of cosine functions.

If $\alpha$ and $\beta$ represent the angles of right triangles, then the cosine of angle alpha is written as $\cos{\alpha}$ and cosine of angle beta is written as $\cos{\beta}$ in mathematics.

Express the Addition of Cosine functions

Write the both cosine functions in a row by displaying a plus sign between them to express the addition of the cosine functions in mathematical form.

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

Expand each cosine function in the expression

Let’s take that $\alpha = a+b$ and $\beta = a-b$. Now, replace the equivalent values of $\alpha$ and $\beta$ in the trigonometric expression.

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

The cosines of compound angles can be expanded by the angle sum and angle difference trigonometric identities of cosine 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

Let’s focus on simplifying the trigonometric expression in the right hand side of the equation by the fundamental mathematical operations.

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

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

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

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

The trigonometric expression is successfully simplified and it expresses the transformation of sum of the cosine functions into product form. However, the product form is in terms of $a$ and $b$. Hence, we should express them in terms of $\alpha$ and $\beta$.

We have assumed that $\alpha = a+b$ and $\beta = a-b$. Now, let’s evaluate $a$ and $b$ in terms of $\alpha$ and $\beta$. They are actually calculated by the fundamental operations of the mathematics.

Add both algebraic equations firstly to evaluate the value of $a$.

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

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

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

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

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

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

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

Now, subtract the equation $\beta = a-b$ from the equation $\alpha = a+b$ to evaluate $b$ in terms of $\alpha$ and $\beta$.

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

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

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

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

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

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

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

Therefore, we have successfully calculated both $a$ and $b$. Now, substitute them in the trigonometric equation $\cos{\alpha}+\cos{\beta}$ $\,=\,$ $2\cos{a}\cos{b}$.

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

Therefore, the sum of the cosine functions is successfully transformed into product form of the trigonometric functions and this equation is called the sum to product identity of cosine functions.

Thus, we can prove the sum to product transformation identity of cosine functions in terms of $x$ and $y$ and also in terms of $C$ and $D$ by following the same procedure.

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