Math Doubts

Difference of sine of compound angles to Product form Transformation

Formula

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

Proof

Two angles can form compound angles in sum and difference form. The subtraction of sines of them can be transformed as the product of cosine and sine of angles. It is called the sum to product form transformation rule as per trigonometry.

Step: 1

Assume, $a$ and $b$ are two angles and the compound angles of them are $a+b$ and $a-b$. The expansion of sine of sum and difference of the angles are written trigonometrically 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

Subtract the expansion of the sine of difference of the angles from the expansion of the sine of sum of the angles.

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

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

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

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

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

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

Verification

Assume $a = 60^\circ$ and $b = 30^\circ$. Substitute these two angles in both sides of the equation and then compare the result.

Step: 1

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

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

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

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

Step: 2

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

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

$\implies 2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $2 \times {\Bigg(\dfrac{1}{2}\Bigg)}^2$

$\implies 2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $2 \times \dfrac{1}{4}$

$\therefore \,\,\,\,\,\, 2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $\dfrac{1}{2}$

Step: 3

Now compare both results to understand the relation between them.

$\therefore \,\,\,\,\,\, \sin(60^\circ+30^\circ) + \sin(60^\circ-30^\circ)$ $=$ $2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $\dfrac{1}{2}$

Other forms

The sum to product form transformation law can also be written in other form.

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

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

General form

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



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