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

The expansion of $\sin{(A-B)}$ trigonometric identity can be derived in geometrical method by constructing a right triangle whose angle is difference of two angles $A$ and $B$.

- $\small \Delta ECD$ is a right triangle and its angle is $\small A$. Draw a line from point $\small C$ to side $\small \overline{DE}$ for dividing the angle $\small A$ as two different angles. If one angle is $\small B$, then the other angle will be $\small A-B$. The line intersects the side $\small \overline{DE}$ at point $\small F$.
- Identify a point ($\small G$) on side $\small \overline{CD}$ and draw a perpendicular line to it but it should intersect the point $\small F$.
- Draw a perpendicular line to side $\small \overline{CE}$ from point $\small G$ and it intersects the side $\small \overline{CF}$ at point $\small H$ and also intersects the side $\small \overline{CE}$ at $\small I$.
- Lastly, draw a perpendicular line to side $\small \overline{IG}$ from point $\small F$ and it intersects the side $\small \overline{IG}$ at point $\small J$.

$A-B$ is an angle of $\Delta ECF$ and it represents the difference of two angles. Now, find the value of sine of difference of angles as a ratio of lengths of the respective sides.

$\overline{EF}$ is opposite side and $\overline{CF}$ is hypotenuse in $\Delta ECF$. The ratio of lengths of them represents the value of $\sin{(A-B)}$ as per trigonometry.

$\sin{(A-B)} \,=\, \dfrac{EF}{CF}$

The sides $\overline{EF}$ and $\overline{JI}$ are parallel lines and equal geometrically. Therefore, the length of side $\overline{EF}$ can be replaced by the length of the side $\overline{JI}$.

$\implies \sin{(A-B)} \,=\, \dfrac{JI}{CF}$

The side $\overline{JF}$ is perpendicularly intersected the side $\overline{GI}$ and it divides the side $\overline{GI}$ as two sides $\overline{GJ}$ and $\overline{JI}$ at point $J$.

$GI = GJ+JI$

$\implies JI = GI-GJ$

Now, replace the length of side $\overline{JI}$ by its equivalent value in the expansion of $\sin{(A-B)}$ identity.

$\implies \sin{(A-B)} \,=\, \dfrac{GI-GJ}{CF}$

$\implies \sin{(A-B)} \,=\, \dfrac{GI}{CF}-\dfrac{GJ}{CF}$

$GI$ is length of opposite side of right triangle $ICG$. So, consider $\Delta ICG$ in which the angle is $A$. Use this information and express length of side $\overline{GI}$ in terms of a trigonometric function.

The relation between angle $A$ and length of opposite side $\overline{GI}$ represents trigonometric ratio sine.

$\sin{A} = \dfrac{GI}{CG}$

$\implies GI = CG \times \sin{A}$

Therefore, replace the length of opposite side $GI$ by its new value.

$\implies \sin{(A-B)} \,=\, \dfrac{CG \times \sin{A}}{CF}-\dfrac{GJ}{CF}$

$\implies \sin{(A-B)} \,=\, \dfrac{CG}{CF} \times \sin{A}-\dfrac{GJ}{CF}$

$CG$ is length of adjacent side and $CF$ is length of hypotenuse of right triangle $GCF$. So, consider $\Delta GCF$ and its angle is $B$.

The relation between adjacent side ($\overline{CG}$), hypotenuse ($\overline{CF}$) and angle $B$ is cosine according to trigonometry. So, express cos of angle $B$ as the ratio of lengths of the sides $\overline{CG}$ to $\overline{CF}$.

$\cos{B} \,=\, \dfrac{CG}{CF}$

Replace the ratio of lengths of sides $\overline{CG}$ to $\overline{CF}$ by $\cos{B}$ in the expansion of $\sin{(A-B)}$ rule.

$\implies \sin{(A-B)} \,=\, \cos{B} \times \sin{A}-\dfrac{GJ}{CF}$

$\implies$ $\sin{(A-B)}$ $\,=\,$ $\sin{A}\cos{B}-\dfrac{GJ}{CF}$

The length of the side $\overline{GJ}$ is part of the right triangle $JGF$ and its angle is unknown. So, it is not possible to derive the expansion of $\sin{(A-B)}$ identity further in trigonometry.

So, the $\angle JGF$ must be found mathematically and it can be done geometrically in four steps by studying the relation between angles of the triangles.

$\Delta ICH$ is a triangle in which an angle is $A-B$ and other angle is a right angle but the third angle is unknown. It can be determined by using sum of angles of a triangle rule.

$\angle ICH + \angle CHI + \angle HIC$ $\,=\,$ $180^°$

$\implies$ $A-B + \angle CHI + 90^°$ $\,=\,$ $180^°$

$\implies$ $\angle CHI$ $\,=\,$ $180^°-90^°-A+B$

$\,\,\, \therefore \,\,\,\,\,\,$ $\angle CHI \,=\, 90^°-A+B$

Therefore, it is derived successfully that the $\angle CHI$ is equal to $90^°-A+B$

$\angle CHI$ and $\angle CHG$ are two angles and the sum of them is a straight angle.

The value of $\angle CHI$ is derived in previous step but the $\angle CHG$ is unknown. However, it can be evaluated by straight angle.

$\angle CHI + \angle CHG \,=\, 180^°$

$\implies$ $90^°-A+B + \angle CHG \,=\, 180^°$

$\implies$ $\angle CHG \,=\, 180^°-90^°+A-B$

$\,\,\, \therefore \,\,\,\,\,\,$ $\angle CHG \,=\, 90^°+A-B$

Thus, it is determined that the $\angle CHG$ is equal to $90^°+A-B$.

$\Delta HCG$ is a triangle in which the sum of angles should be equal to $180^°$ as per sum of angles in a triangle rule. An angle of this triangle is $B$ and other angle is evaluated in the above step.

$\angle HCG + \angle CGH + \angle GHC$ $\,=\,$ $180^°$

$\implies$ $\angle HCG + \angle CGH + \angle CHG$ $\,=\,$ $180^°$

$\implies$ $B + \angle CGH + 90^°+A-B$ $\,=\,$ $180^°$

$\implies$ $\angle CGH + 90^°+A-B+B$ $\,=\,$ $180^°$

$\implies$ $\require{cancel} \angle CGH + 90^°+A-\cancel{B}+\cancel{B}$ $\,=\,$ $180^°$

$\implies$ $\angle CGH + 90^°+A$ $\,=\,$ $180^°$

$\implies$ $\angle CGH$ $\,=\,$ $180^°-90^°-A$

$\,\,\, \therefore \,\,\,\,\,\,$ $\angle CGH = 90^°-A$

The side $\overline{GF}$ is a perpendicular line to the side $\overline{CD}$. Therefore, the $\angle FGC$ is a right angle. It says that the sum of the angles $\angle FGJ$ and $\angle JGC$ (or $\angle HGC$) is equal to $90^°$.

$\angle HGC + \angle HGF = 90^°$

$\implies$ $\angle CGH + \angle HGF = 90^°$

$\implies$ $90^°-A + \angle HGF = 90^°$

$\implies$ $\angle HGF = 90^°-90^°+A$

$\implies$ $\require{cancel} \angle HGF = \cancel{90^°}-\cancel{90^°}+A$

$\,\,\, \therefore \,\,\,\,\,\,$ $\angle HGF = A$

Geometrically, the $\angle HGF$ and $\angle JGF$ are same. Therefore, the angle of $\Delta JGF$ is $A$. Actually, the angle of right triangle $ECD$ is $A$. It means the two triangles are congruent.

Comeback to deriving expansion of $\sin{(A-B)}$ formula.

$\implies$ $\sin{(A-B)}$ $\,=\,$ $\sin{A}\cos{B}-\dfrac{GJ}{CF}$

$\overline{GJ}$ is a side of $\Delta JGF$ and it is derived mathematically that the angle of this triangle is $A$. Now, express length of the side $\overline{GJ}$ in the form a trigonometric function.

$\cos{A} = \dfrac{GJ}{GF}$

$\implies GJ = GF \times \cos{A}$

Replace the length of the side $\overline{GJ}$ by its equal value in the expansion of sin of difference of two angles law.

$\implies$ $\sin{(A-B)}$ $\,=\,$ $\sin{A}\cos{B}-\dfrac{GF \times \cos{A}}{CF}$

$\implies$ $\sin{(A-B)}$ $\,=\,$ $\sin{A}\cos{B}-\dfrac{GF}{CF} \times \cos{A}$

$\overline{GF}$ is opposite side and $\overline{CF}$ is hypotenuse of right triangle $FCG$ and its angle is $B$. The ratio of lengths of them represents sine of angle $B$ in trigonometry.

$\sin{B} \,=\, \dfrac{GF}{CF}$

The ratio of lengths of them can be replaced by trigonometric function $\sin{B}$ in the expansion of sin of difference of angles identity.

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

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

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

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