Sum of sine of compound angles to Product form Transformation

Formula

$\large \sin (a+b) + \sin (a-b) =$ $\large 2.\sin a .\cos b$

Proof

When two compound angles are formed by the addition and difference of two angles, the summation of sine of those two compound angles can be transformed in product form and it is equal to twice the product of sine of one angle and cosine of other angle.

The trigonometric identity is called sum to product form transformation rule of the trigonometry.

Step: 1

For example, $a$ and $b$ are two angles. $a+b$ and $a-b$ are two compound angles and expansions of sines of them are expressed as follows.

$(1). \,\,\,\,\,\,$ $\sin(a+b)$ $=$ $\sin a . \cos b$ $+$ $\cos a . \sin b$

$(2). \,\,\,\,\,\,$ $\sin(a-b)$ $=$ $\sin a . \cos b$ $-$ $\cos a . \sin b$

Step: 2

Add both trigonometric equations to obtain sum of them.

$\implies \sin(a+b) + \sin(a-b)$ $=$ $(\sin a . \cos b$ $+$ $\cos a . \sin b)$ $+$ $(\sin a . \cos b$ $-$ $\cos a . \sin b)$

$\implies \sin(a+b) + \sin(a-b)$ $=$ $\sin a . \cos b$ $+$ $\cos a . \sin b$ $+$ $\sin a . \cos b$ $-$ $\cos a . \sin b$

$\implies \sin(a+b) + \sin(a-b)$ $=$ $\sin a . \cos b$ $+$ $\sin a . \cos b$ $+$ $\cos a . \sin b$ $-$ $\cos a . \sin b$

$\implies \sin(a+b) + \sin(a-b)$ $=$ $2 \sin a . \cos b$ $+$ $\require{cancel} \cancel{\cos a . \sin b}$ $-$ $\require{cancel} \cancel{\cos a . \sin b}$

$\therefore \,\,\,\,\,\, \sin(a+b) + \sin(a-b)$ $=$ $2 \sin a . \cos b$

It is transformed that the summation of sine of sum of two angles and sine of difference of two angles is equal to twice the product of sine of one angle and cosine of second angle.

Verification

Take $a = 60^\circ$ and $b = 30^\circ$. Substitute them in both sides of the equation.

Step: 1

$\sin(a+b) + \sin(a-b)$ $=$ $\sin(60^\circ+30^\circ) + \sin(60^\circ-30^\circ)$

$\implies \sin(60^\circ+30^\circ) + \sin(60^\circ-30^\circ)$ $=$ $\sin(90^\circ) + \sin(30^\circ)$

$\implies \sin(60^\circ+30^\circ) + \sin(60^\circ-30^\circ)$ $=$ $1+\dfrac{1}{2}$

$\therefore \,\,\,\,\,\, \sin(60^\circ+30^\circ) + \sin(60^\circ-30^\circ)$ $=$ $\dfrac{3}{2}$

Step: 2

$2 \sin a . \cos b = 2 \times \sin(60^\circ) \times \cos(30^\circ)$

$\implies 2 \times \sin(60^\circ) \times \cos(30^\circ)$ $=$ $2 \times \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2}$

$\implies 2 \times \sin(60^\circ) \times \cos(30^\circ)$ $=$ $2 \times {\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)}^2$

$\implies 2 \times \sin(60^\circ) \times \cos(30^\circ)$ $=$ $2 \times \dfrac{3}{4}$

$\therefore \,\,\,\,\,\, 2 \times \sin(60^\circ) \times \cos(30^\circ)$ $=$ $\dfrac{3}{2}$

Step: 3

Compare both results.

$\therefore \,\,\,\,\,\, \sin(60^\circ+30^\circ) + \sin(60^\circ-30^\circ)$ $=$ $2 \times \sin(60^\circ) \times \cos(30^\circ)$ $=$ $\dfrac{3}{2}$

Other forms

The sum to product form transformation of sine terms can also be written in other forms in general.

$\large \sin (x+y) + \sin (x-y) =$ $\large 2.\sin x .\cos y$

$\large \sin (\alpha+\beta) + \sin (\alpha-\beta) =$ $\large 2.\sin \alpha .\cos \beta$

General form

$\sin (sum \, of \, two \, angles)$ $+$ $\sin (difference \, of \, two \, angles)$ $=$ $2$ $\times$ $\sin \, of \, first \, angle$ $\times$ $\cos of \, second \, angle$

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