# $\sin{(A+B)}$ formula Proof

## Formula

$\sin{(A+B)}$ $\,=\,$ $\sin{A}\cos{B}$ $+$ $\cos{A}\sin{B}$

### Construction of sum of angles triangle

The expansion of $\sin{(A+B)}$ identity is actually derived in geometrical approach by constructing a right angled triangle, in which its angle is divided as two different angles.

1. $\small \Delta EDF$ is a right angled triangle and draw line from $\small D$ to divide the angle of $\small \Delta EDF$ as two different angles $\small A$ and $\small B$. The line intersects side $\small \overline{EF}$ at point $\small G$.
2. Draw a line from point $\small G$ to side $\small \overline{FD}$ but it should be perpendicular to side $\small \overline{DG}$ and the line intersects the side $\small \overline{FD}$ at point $\small H$.
3. Draw a perpendicular to side $\small \overline{ED}$ from point $\small H$ and it intersects the side $\small \overline{DG}$ at point $\small K$ and also intersects the side $\small \overline{ED}$ at point $\small J$.
4. Finally, draw a perpendicular line to side $\small \overline{HJ}$ from $\small G$ and it intersects the side $\small \overline{HJ}$ at point $\small K$. Similarly, draw a perpendicular line from $\small I$ to side $\small \overline{EF}$ and intersects the side $\small \overline{EF}$ at point $\small L$.

### Express Sin of sum of angles in equal ratio form

$A+B$ is angle of the $\Delta JDH$ and it is useful to express sin of sum of angles as a ratio of lengths of the opposite side to hypotenuse.

$(1) \,\,\,\,\,\,$ $\overline{HJ}$ is opposite side

$(2) \,\,\,\,\,\,$ $\overline{DH}$ is hypotenuse

The lengths of them are written as $HJ$ and $DH$ respectively in mathematics.

$\sin{(A+B)} = \dfrac{HJ}{DH}$

### Express length of side in alternative form

The side $\overline{KG}$ is perpendicularly intersected the side $\overline{HJ}$ at point $K$. Therefore, the length of the side $\overline{HJ}$ can be written as sum of the lengths of the sides $\overline{HK}$ and $\overline{KJ}$.

$HJ = HK+KJ$

Now, replace the length of the side $\overline{HJ}$ in the equivalent expansion of $\sin{(A+B)}$ law.

$\implies$ $\sin{(A+B)} = \dfrac{HK+KJ}{DH}$

$\implies$ $\sin{(A+B)} = \dfrac{HK}{DH}+\dfrac{KJ}{DH}$

$\overline{KJ}$ and $\overline{GE}$ are parallel lines and also equal because of the perpendicular intersection of another set of parallel lines $\overline{KG}$ and $\overline{JE}$.

$KJ = GE$

Now, substitute the length of the side $\overline{KJ}$ by the length of the side $\overline{GE}$.

$\implies$ $\sin{(A+B)} = \dfrac{HK}{DH}+\dfrac{GE}{DH}$

### Write each side in a trigonometric function form

$\overline{GE}$ is opposite side of right angled triangle $EDG$ and $A$ is an angle of that triangle. So, write the length of the side $\overline{GE}$ in trigonometric function form.

$\sin{A} = \dfrac{GE}{DG}$

$\implies GE = DG\sin{A}$

Now, replace the length of the opposite side $\overline{GE}$ by its equivalent value.

$\sin{(A+B)} = \dfrac{HK}{DH}+\dfrac{DG\sin{A}}{DH}$

$\overline{HK}$ is a side of right angled triangle $KHG$ but its angle is unknown. So, it is not possible to express the length of side $\overline{HK}$ in trigonometric form. This problem can be overcome by some geometrical analysis.

$\Delta EDG$ and $\Delta LIG$ are right angled triangles and moreover, they are similar triangles. Therefore, their corresponding angles should be equal geometrically. In this case, $\angle EDG$ and $\angle LIG$ are corresponding angles but $\angle EDG = A$.

$\angle LIG = \angle EDG = A$

$\overline{KG}$ and $\overline{IL}$ are parallel lines and $\overline{IG}$ is their transversal. $\angle LIG$ and $\angle IGK$ are interior alternate angles and they are equal geometrically.

$\angle IGK = \angle LIG = A$

The side $\overline{HG}$ is drawn perpendicularly to the side $\overline{DG}$. So, the sum of $\angle IGK$ and $\angle KGH$ should be $90^\circ$.

$\angle IGK + \angle KGH = 90^\circ$
$\implies A + \angle KGH = 90^\circ$
$\implies \angle KGH = 90^\circ-A$

$\Delta KHG$ is a right angled triangle. In this triangle, $\angle HKG = 90^\circ$, $\angle KGH = 90^\circ-A$ but $\angle GHK$ is unknown. However, it can be calculated by using sum of angles in a triangle rule.

$\angle GHK + \angle KGH + \angle HKG$ $\,=\,$ $180^\circ$
$\implies \angle GHK + 90^\circ-A + 90^\circ$ $\,=\,$ $180^\circ$
$\implies \angle GHK + 180^\circ-A$ $\,=\,$ $180^\circ$
$\implies \angle GHK$ $\,=\,$ $180^\circ-180^\circ+A$
$\implies \angle GHK \,=\, A$

$\cos{A} = \dfrac{HK}{HG}$

$\implies HK = HG\cos{A}$

Replace the length of the side $\overline{HK}$ by its equivalent value.

$\implies \sin{(A+B)}$ $\,=\,$ $\dfrac{HG\cos{A}}{DH}+\dfrac{DG\sin{A}}{DH}$

$\implies \sin{(A+B)}$ $\,=\,$ $\dfrac{HG}{DH} \times \cos{A} +\dfrac{DG}{DH} \times \sin{A}$

### Express each ratio in trigonometric function form

In the expansion of $\sin(A+B)$ rule, $HG$, $DG$ and $DH$ are remaining factors in the trigonometric expression. Actually, the three of them are lengths of the sides of the $\Delta GDH$ and $B$ is the angle of this right angled triangle. Now express each ratio of lengths of the sides by its respective trigonometric function.

$\sin{B} \,=\, \dfrac{HG}{DH}$

$\cos{B} \,=\, \dfrac{DG}{DH}$

Replace the ratio of lengths of the sides by the corresponding trigonometric function in the expansion of $\sin{(A+B)}$ rule.

$\implies \sin{(A+B)}$ $\,=\,$ $\sin{B}\times\cos{A}+\cos{B}\times\sin{A}$

$\implies \sin{(A+B)}$ $\,=\,$ $\sin{B}\cos{A}+\cos{B}\sin{A}$

$\implies \sin{(A+B)}$ $\,=\,$ $\cos{A}\sin{B}+\sin{A}\cos{B}$

$\,\,\, \therefore \,\,\,\,\,\, \sin{(A+B)}$ $\,=\,$ $\sin{A}\cos{B}+\cos{A}\sin{B}$

This trigonometric identity is used to expand sin of sum of two angles in terms of sine and cosine of those two angles.