Math Doubts

Difference to Product identity of Sin functions

Formula

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

The transformation of difference of two sin functions into product form is called the difference to product identity of sin functions.

Introduction

Let $\alpha$ and $\beta$ represent two angles of two right triangles, the sine functions with the same angles are written in mathematical form as $\sin{\alpha}$ and $\sin{\beta}$. In trigonometry, the sine functions participate in subtraction in some cases and the difference of them is expressed in mathematical form as follows.

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

The subtraction of the sine functions can be transformed into product form for simplifying it.

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

Popular forms

The difference to product transformation rule of sin functions is also popularly written in two other forms.

$(1) \,\,\,\,\,\,$ $\sin{x}-\sin{y}$ $\,=\,$ $2\cos{\Big(\dfrac{x+y}{2}\Big)}\sin{\Big(\dfrac{x-y}{2}\Big)}$

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

In this way, you can express the difference to product formula of sine functions in terms of any two angles.

Proof

Learn how to derive the difference to product transformation identity of sin functions in trigonometry.

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