Math Doubts

Proof of $\sin{(A-B)}$ formula in Geometric method

Formula

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

Construction of a triangle with difference of angles

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 $B$.

construction of difference of angles triangle for sin(a-b) formula
  1. $\small \Delta ECD$ 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 $\small F$.
  2. Identify a point ($\small G$) on side $\small \overline{CD}$ and draw a perpendicular line to it but it should intersect the point $\small F$.
  3. 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 $\small I$.
  4. Lastly, draw a perpendicular line to side $\small \overline{IG}$ from point $\small F$ and it intersects the side $\small \overline{IG}$ at point $\small J$.

Write sin of difference of angles as a ratio of sides

$A-B$ 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.

a+b triangle for sin(a+b) formula

$\overline{EF}$ 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 $\overline{JI}$.

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

a+b triangle for sin(a+b) formula

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.

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

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

Express lengths of sides in alternative form

$GI$ 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.

angle a right triangle for deriving sin(a-b) formula expansion

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 $B$.

angle b right triangle for deriving sin(a-b) formula expansion

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.

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

Find the angle of Right triangle

finding unknown of right triangle for sin(a-b) rule

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

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.

Find unknown angle of triangle

evaluating unknown third angle for sin(a-b) formula

$\Delta ICH$ 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 $90^°-A+B$

Find part angle of the straight angle

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

unknown part of straight angle for sin(a-b) rule

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 $90^°+A-B$.

Find unknown angle of triangle

evaluating unknown third angle for sin(a-b) law

$\Delta HCG$ 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.

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

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

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

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

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

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

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

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

Find angle of the right triangle

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 $90^°$.

finding unknown of right triangle for sin(a-b) rule

$\angle HGC + \angle HGF = 90^°$

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

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

$\implies$ $\angle HGF = 90^°-90^°+A$

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

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

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.

Write each side in a trigonometric function form

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

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

congruent angle a triangle for deriving expansion of sin(a-b) formula

$\overline{GJ}$ 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.

angle b right triangle for deriving sin(a-b) formula expansion

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

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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