The angle sum cosine identity is used as a formula to expanded cosine of sum of two angles. For example, $\cos{(A+B)}$, $\cos{(x+y)}$, $\cos{(\alpha+\beta)}$, and so on. Here, you learn how cos of sum of two angles formula is derived in geometric method.

The $\Delta EDF$ is a right triangle and the angle of this triangle is divided as two angles. It’s useful in deriving the cosine of sum of two angles trigonometric identity geometrically.

$(1). \,\,\,$ Firstly, draw a straight line to side $\overline{EF}$ from point $D$ for dividing the $\angle EDF$ as two angles $x$ and $y$, and it intersects the side $\overline{EF}$ at point $G$.

$(2). \,\,\,$ Draw a straight line to side $\overline{DE}$ from point $G$ but it should be perpendicular to the side $\overline{DG}$. It means $\overline{DG} \perp \overline{HG}$

$(3). \,\,\,$ Similarly, draw a perpendicular line to side $\overline{DF}$ from point $H$, but it intersects the side $\overline{DG}$ at point $I$ and also perpendicularly intersects the side $\overline{DF}$ at point $J$.

$(4). \,\,\,$ Lastly, draw a perpendicular line to side $\overline{HJ}$ from point $G$ and it intersects the side $\overline{HJ}$ at point $K$.

Thus, the $\Delta GDF$ having $x$ as its angle, the $\angle EDG$ having $y$ as its angle and $\angle KHG$ with unknown angle are formed geometrically.

Now, let’s start the geometrical approach to derive the angle sum cosine identity in the form of $\cos{(x+y)}$ formula in trigonometry on the basis of the above constructed triangle.

The $\angle JDH$ is $x+y$ in the $\Delta JDH$ and write the cos of compound angle $x+y$ in its ratio from.

$\cos{(x+y)}$ $\,=\,$ $\dfrac{DJ}{DH}$

The side $\overline{HJ}$ divides the side $\overline{DF}$ as two parts. So, the sum of lengths of the sides $\overline{DJ}$ and $\overline{JF}$ is equal to the length of the side $\overline{DF}$.

$DF \,=\, DJ+JF$

$\implies DJ \,=\, DF-JF$

Substitute the length of the side $\overline{DJ}$ by its equivalent value in the expansion of $\cos{(x+y)}$.

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\dfrac{DF-JF}{DH}$

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\dfrac{DF}{DH}$ $\,-\,$ $\dfrac{JF}{DH}$

The lengths of sides $\overline{JF}$ and $\overline{KG}$ are equal because they both are parallel lines. Therefore, $JF = KG$

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\dfrac{DF}{DH}$ $\,-\,$ $\dfrac{KG}{DH}$

The side $\overline{DF}$ is adjacent side of $\Delta FDG$ and angle of this triangle is $x$. It is possible to express the length of the side $\overline{DF}$ in a trigonometric function.

$\cos{x} \,=\, \dfrac{DF}{DG}$

$\implies DF \,=\, {DG}\cos{x}$

Replace the length of the side $\overline{DF}$ by its equivalent value in the $\cos{(x+y)}$ expansion.

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

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x} \times \dfrac{DG}{DH}$ $\,-\,$ $\dfrac{KG}{DH}$

$DG$ and $DH$ are lengths of the sides $\overline{DG}$ and $\overline{DH}$ respectively. They are sides of the $\Delta GDH$ and its angle is $y$. The ratio between them can be represented by cos of angle $y$.

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

The ratio of $DG$ to $DH$ can be replaced by $\cos{y}$ in the expansion of $\cos{(x+y)}$.

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x} \times \cos{y}$ $\,-\,$ $\dfrac{KG}{DH}$

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\dfrac{KG}{DH}$

Similarly, the ratio of $KG$ to $DH$ can also be expressed in terms of trigonometric functions but it is not possible at this time because it’s the side of the $\Delta KHG$ and its angle is unknown.

Now, it’s time to find the $\angle KHG$ and it is useful to us in expressing the length of the side $\overline{KG}$ in a trigonometric function. The sides $\overline{KG}$ and $\overline{DF}$ are parallel lines and their transversal is $\overline{DG}$. The $\angle FDG$ and $\angle KGD$ are alternate interior angles and they are equal geometrically.

$\angle KGD \,=\, \angle FDG$ but $\angle FDG = x$

$\therefore \,\,\,\,\,\, \angle KGD \,=\, x$

Actually $\overline{DG} \perp \overline{HG}$. So, $\angle DGH = 90^°$. The side $\overline{KG}$ splits the $\angle DGH$ as two angles $\angle KGD$ and $\angle KGH$.

$\angle DGH \,=\, \angle KGD + \angle KGH$

$\implies 90^° = x+\angle KGH$

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

Two angles are known in $\Delta KHG$ and its third angle $\angle KHG$ can be calculated from them by using sum of angles of a triangle rule.

$\angle KHG + \angle HKG + \angle KGH$ $\,=\,$ $180^°$

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

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

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

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

$\,\,\, \therefore \,\,\,\,\,\,$ $\angle KHG \,=\, x$

It is derived that $\angle KHG = x$ but the $\angle FDG = x$ and it clears that they both are congruent. Therefore, $\Delta FDG$ and $\Delta KHG$ are similar triangles.

Let’s continue deriving the expansion of the angle sum cosine identity.

$\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\dfrac{KG}{DH}$

$KG$ is length of the side $\overline{KG}$ in $\Delta KHG$ and it can be written in terms of trigonometric function.

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

$\implies KG \,=\, {HG}\sin{x}$

Therefore, the length of the side $\overline{KG}$ by its equivalent value in the expansion of cos of compound angle.

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\dfrac{{HG}\sin{x}}{DH}$

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\sin{x} \times \dfrac{HG}{DH}$

$HG$ and $DH$ are lengths of the sides $\overline{HG}$ and $\overline{DH}$ in $\Delta GDH$. The angle of $\Delta GDH$ is $y$. So, the ratio between them can be represented by the trigonometric function $\sin{y}$.

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

Therefore, the ratio of $HG$ to $DH$ can be replaced by $\sin{y}$ in the expansion of the $\cos{(x+y)}$ function.

$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\sin{x} \times \sin{y}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\sin{x}\sin{y}$

Therefore, it is proved that the cos of sum of two angles can be expanded as the subtraction of product of sines of angles from product of cosines of the angles.

Latest Math Topics

Apr 18, 2022

Apr 14, 2022

Apr 05, 2022

Mar 18, 2022

Mar 05, 2022

Latest Math Problems

Apr 06, 2022

Mar 22, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved