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

The expansion of $\cos{(A+B)}$ formula is originally derived in geometric method by constructing a right triangle whose angle is sum of two angles.

- $\small \Delta FDE$ is a right triangle. Draw a straight line from point $\small D$ and it divides the angle as two angles $\small A$ and $\small B$ and intersects the side $\small \overline{FE}$ at point $\small G$.
- Later, draw a line from point $\small G$ to side $\small \overline{DF}$ but it should be perpendicular to the side $\small \overline{DG}$ and it interests the side $\small \overline{DF}$ at point $\small H$.
- Draw a perpendicular line to side $\small \overline{DE}$ from point $\small H$. It intersects the side $\small \overline{DG}$ at point $\small I$ firstly and then side $\small \overline{DE}$ at point $\small J$.
- Lastly, draw a perpendicular line from point $\small G$ to side $\small \overline{HJ}$ and also draw a perpendicular line from point $\small I$ to side $\small \overline{FE}$.

The angle of $\Delta HDJ$ is a sum of the angles $A$ and $B$. So, write cos of angle $A+B$ in its ratio form for deriving expansion of cosine of sum of two angles.

- $\small \overline{DJ}$ is Adjacent side (or) Base
- $\small \overline{DH}$ is hypotenuse

Express, cosine of sum of the angles in ratio form of the sides by considering right triangle $HDJ$.

$\cos{(A+B)} \,=\, \dfrac{DJ}{DH}$

The side $\overline{HJ}$ perpendicularly divides the side $\overline{DE}$ at point $J$. So, the sum of the lengths of the sides $\overline{DJ}$ and $\overline{JE}$ is equal to length of the side $\overline{DE}$.

$DE = DJ+JE$

$\implies DJ = DE-JE$

The length of the side $\overline{DJ}$ in expansion of $\cos{(A+B)}$ can be replaced by its equal value.

$\implies$ $\cos{(A+B)}$ $\,=\,$ $\dfrac{DE-JE}{DH}$

The perpendicular intersections of sides $\overline{DE}$ and $\overline{KG}$ with the sides $\overline{HJ}$ and $\overline{EF}$ form a rectangle. It is the main reason for the equality of the sides $\overline{JE}$ and $\overline{KG}$.

$JE = KG$

Replace the length of the side $\overline{JE}$ in the expansion of $\cos{(A+B)}$ identity.

$\implies$ $\cos{(A+B)}$ $\,=\,$ $\dfrac{DE-KG}{DH}$

$\implies$ $\cos{(A+B)}$ $\,=\,$ $\dfrac{DE}{DH}-\dfrac{KG}{DH}$

The length of the side $\overline{DE}$ is part of the right triangle $EDG$ and its angle is $A$. Now, write length of the side $\overline{DE}$ in another form by considering $\Delta EDG$.

In $\Delta EDG$, $A$ is angle and $\overline{DE}$ is adjacent side. So, The relation between them is represented by cosine. So, express length of side ${DE}$ in terms of cos of angle $A$.

$\cos{A} = \dfrac{DE}{DG}$

$\implies DE = DG\cos{A}$

Substitute the length of the side $\overline{DE}$ in the expansion of $\cos{(A+B)}$ rule by its new equal value.

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

The expansion of cos of sum of angles is derived in the previous step as follows.

$\cos{(A+B)} \,=\, \dfrac{DG\cos{A}}{DH}-\dfrac{KG}{DH}$

$KG$ is length of the opposite side ($\overline{KG}$) of right triangle $GHK$ and the angle of this triangle is unknown. So, it is not possible to express the length of side $\overline{KG}$ in terms of a trigonometric function but it is possible by finding the angle of this triangle.

The $\angle GHK$ is about to find in three steps geometrically.

$\Delta EDG$ is a right triangle and its angle is $A$. $\Delta LIG$ is another right triangle and its angle is unknown.

Geometrically, the triangles $\Delta EDG$ and $\Delta LIG$ are similar.

$\Delta EDG \, \sim \, \Delta LIG$

Therefore, the corresponding angles $\angle LIG$ and $\angle EDG$ should be equal.

$\implies \angle LIG = \angle EDG$

$\,\,\, \therefore \,\,\,\,\,\, \angle LIG = A$

Therefore, the $\angle LIG$ is equal to $A$.

$\overline{DG}$ is a transversal of parallel lines $\overline{IL}$ and $\overline{KG}$. The $\angle LIG$ and $\angle KGI$ are known as interior alternate angles.

It is proved that the alternate interior angles formed by the parallel lines and their transversal are equal geometrically.

$\implies \angle KGI \,=\, \angle LIG$

$\,\,\, \therefore \,\,\,\,\,\, \angle KGI = A$

Therefore, the $\angle KGI$ is equal to $A$.

The side $\overline{HG}$ is a perpendicular line to side $\overline{DG}$. So, $\angle DGH = 90^°$.

The intersection of side $\overline{KG}$ divides the $\angle DGH$ as two angles and they are $\angle DGK$ and $\angle KGH$. Therefore, the sum of them is equal to $\angle DGH$.

$\angle DGK + \angle KGH = \angle DGH$

$\implies \angle DGK + \angle KGH = 90^°$

Geometrically, $\angle DGK = \angle KGI$ but $\angle KGI = A$ as per the above step.

$\angle KGI + \angle KGH = 90^°$

$\implies A + \angle KGH = 90^°$

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

$\Delta KHG$ is a right triangle and its angle is unknown but the other two angles are known. So, the angle this triangle can be evaluated by using sum of interior angles in a triangle rule.

$\angle KHG + \angle HGK + \angle GKH \,=\, 180^°$

$\implies \angle KHG + \angle KGH + \angle GKH \,=\, 180^°$

$\angle GKH$ is a right angle and it is derived in the previous step that $\angle KGH = 90^°-A$. Substitute them and get the value of $\angle KHG$ geometrically.

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

$\implies \angle KHG + 180^°-A \,=\, 180^°$

$\implies \angle KHG \,=\, 180^°-180^°+A$

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

$\,\,\, \therefore \,\,\,\,\,\, \angle KHG \,=\, A$

Therefore, the angle of $\Delta KHG$ is $A$.

Return to deriving the expansion of $\cos{(A+B)}$ formula.

$\cos{(A+B)}$ $\,=\,$ $\dfrac{DG\cos{A}}{DH}-\dfrac{KG}{DH}$

The angle of $\Delta KHG$ is determined in the above step. So, the length of the side can be expressed in the form a trigonometric function.

$KG$ is opposite site of the $\Delta KHG$ and the angle of this triangle is $A$. So, the relationship between them can be represented by sine function.

$\sin{A} \,=\, \dfrac{KG}{HG}$

$\implies KG = HG \times \sin{A}$

Now, replace the length of the side KG in the expansion of cos of sum of two angles.

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

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

$DG$, $HG$ and $DH$ are lengths of three sides of the right angled triangle ($\Delta HDG$) whose angle is $B$. Now, express value of each ratio of the sides in the form of corresponding trigonometric function.

$(1) \,\,\,\,\,\,$ $\sin{B} \,=\, \dfrac{HG}{DH}$

$(2) \,\,\,\,\,\,$ $\cos{B} \,=\, \dfrac{DG}{DH}$

Now, substitute the ratios of lengths of the sides by their associated trigonometric functions in the expansion of $\cos{(A+B)}$ identity.

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

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

Therefore, it is proved that cos of sum of two angles is expanded as the subtraction of product of sin of angles from product of cos of angles.

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