# Deriving the trigonometric addition formulae using area and cosine rule

###### Abstract

The trigonometric addition formulae are very important and useful in Mathematics. They can be derived from geometric proof, relations from analytical geometry, Euler formula or relations from vectorial analysis. In this work, we prove the sine addition formula  by considering area. Using cosine rule, we prove the cosine addition formula. Our proof is simpler because it requires the knowledge of area of triangles and cosine rule which are easily understood by students. From the addition formulae, we obtain the difference formulae by solving simultaneous equations.

Keywords
Trigonometric addition and difference formulae, area of triangles, cosine rule

Deriving the trigonometric addition formulae using area and cosine rule

Deriving the trigonometric addition formulae using area and cosine rule

## 1 Introduction

The sine and cosine addition and difference formulae are given by

 $\sin{(\alpha\pm\beta)}=\sin{\alpha}\cos{\beta}\pm\sin{\beta}\cos{\alpha}$ (1)

and

 $\cos{(\alpha\pm\beta)}=\cos{\alpha}\cos{\beta}\mp\sin{\alpha}\sin{\beta},$ (2)

respectively. The sine and cosine addition and diference formulae can be obtained from geometric proofs , Euler’s formula, analytical geometry and vectorial analysis . Feldman uses the cosine rule and Pythagoras theorem  to get the cosine difference formula. In this work, we consider the area of triangles to obtain the sine addition formula since the area of any triangle depends on the sine of the angle. Then, using the cosine rule twice, we obtain the cosine addition formula. Finally, we obtain the trigonometric difference formulae from addition formulae by solving two simultaneous equations.

## 2 Preliminaries

We use the important formulas and identities:
From Fig. (1), we have

 $\text{area of \triangle XYZ}=\frac{1}{2}xy\sin{\theta}$

and the cosine rule gives

 $z^{2}=x^{2}+y^{2}-2xy\cos{\theta}.$

The simple trigonometric identities are given by

 $\displaystyle\sin{(180^{\circ}-\theta)}=\sin{\theta},$ (3) $\displaystyle\cos{(180^{\circ}-\theta)}=-\cos{\theta},$ (4) $\displaystyle\sin^{2}{\theta}+\cos^{2}{\theta}=1.$ (5)

## 3 Derivation of Addition Trigonometric Formulae

Fig. (2) shows a quadrilateral ABCD which consists of two right-angled triangles $ABC$ and $ACD$. In $\triangle\ ABC$, $C\widehat{A}B=\alpha$, $AC=s$. By simple trigonometry  , we have

 $\begin{array}[]{cc}AB=s\ \cos{\alpha},&BC=s\ \sin{\alpha}.\end{array}$ (6)

In $\triangle\ ACD$, $C\widehat{A}D=\beta$, $AD=h$. We have the following trigonometric equations:

 $\begin{array}[]{cc}\cos{\beta}=\displaystyle{\frac{s}{h}},&CD=h\ \sin{\beta}.% \end{array}$ (7)

We also note that $B\widehat{C}D=180^{\circ}-\alpha.$
We can express the area of quadrilateral ABCD as:

 $\displaystyle\text{Area of}\ \triangle\ ABC+\text{Area of}\ \triangle\ ACD=% \text{Area of}\ \triangle\ ABD+\text{Area of}\ \triangle\ BCD$ $\displaystyle\frac{1}{2}\ (AB)\ (BC)+\frac{1}{2}\ (AC)\ (CD)=\frac{1}{2}\ (AB)% \ h\ \sin{(\alpha+\beta)}$ $\displaystyle +% \frac{1}{2}\ (CD)\ (BC)\ \sin{(180^{\circ}-\alpha)}.$ (8)

Substituting eqs. (3), (6) and (7) into eq. (3), we have, after simplifications,

 $\frac{1}{2}s^{2}\sin{\alpha}\cos{\alpha}+\frac{1}{2}\ hs\sin{\beta}=\frac{1}{2% }hs\cos{\alpha}\sin{(\alpha+\beta)}+\frac{1}{2}hs\sin{\beta}\sin^{2}{\alpha}.$ (9)

Dividing eq. (9) by $\displaystyle{\frac{1}{2}hs},$ we obtain

 $\frac{s}{h}\sin{\alpha}\cos{\alpha}+\sin{\beta}=\cos{\alpha}\sin{(\alpha+\beta% )}+\sin{\beta}\sin^{2}{\alpha}.$

Simplifying and using eq. (5), we have

 $\displaystyle\cos{\alpha}\sin{(\alpha+\beta)}$ $\displaystyle=$ $\displaystyle\frac{s}{h}\sin{\alpha}\cos{\alpha}+\sin{\beta}(1-\sin^{2}{\alpha% }),$ (10) $\displaystyle=$ $\displaystyle\frac{s}{h}\sin{\alpha}\cos{\alpha}+\sin{\beta}\cos^{2}{\alpha}.$

Dividing eq. (10) by $\cos{\alpha}$ and using eq. (7), we finally obtain the sine addition formula given by eq. (1) with the $+$ sign.
We next consider $\triangle\ BCD$. By using cosine rule and eqs. (4), (6) and (7), we have

 $\displaystyle BD^{2}$ $\displaystyle=$ $\displaystyle CD^{2}+BC^{2}-2(CD)(BC)\cos{(180^{\circ}-\alpha)},$ (11) $\displaystyle=$ $\displaystyle h^{2}\sin^{2}{\beta}+s^{2}\sin^{2}{\alpha}+2hs\sin{\beta}\sin{% \alpha}\cos{\alpha}.$

Using cosine rule in $\triangle\ ABD$ and eq. (11), we obtain

 $\displaystyle\cos{(\alpha+\beta)}$ $\displaystyle=$ $\displaystyle\frac{h^{2}+AB^{2}-BD^{2}}{2h(AB)}$ (12) $\displaystyle=$ $\displaystyle\frac{h^{2}(1-\sin^{2}{\beta})+s^{2}(\cos^{2}{\alpha}-\sin^{2}{% \alpha})-2hs\sin{\beta}\sin{\alpha}\cos{\alpha}}{2hs\cos{\alpha}}.$

Using eq. (5) and $s^{2}=h^{2}\cos^{2}{\beta}$, eq. (12) reduces to

 $\displaystyle\cos{(\alpha+\beta)}$ $\displaystyle=$ $\displaystyle\frac{h^{2}\cos^{2}{\beta}+s^{2}(\cos^{2}{\alpha}-\sin^{2}{\alpha% })-2hs\sin{\beta}\sin{\alpha}\cos{\alpha}}{2hs\cos{\alpha}}$ (13) $\displaystyle=$ $\displaystyle\frac{s^{2}(1+\cos^{2}{\alpha}-\sin^{2}{\alpha})-2hs\sin{\beta}% \sin{\alpha}\cos{\alpha}}{2hs\cos{\alpha}}$ $\displaystyle=$ $\displaystyle\frac{2s^{2}\cos^{2}{\alpha}-2hs\sin{\beta}\sin{\alpha}\cos{% \alpha}}{2hs\cos{\alpha}}$ $\displaystyle=$ $\displaystyle\frac{s}{h}\cos{\alpha}-\sin{\alpha}\sin{\beta}.$

Substituting eq. (7) into eq. (13), we finally obtain cosine addition formula given by eq. (2) with the $+$ sign. We point out the proof of the cosine addition formula is a generalization to that of Feldman who used $s=h=1.$

## 4 Derivation of Trigonometric Difference Formulae

The difference formulas can be obtained by replacing $\beta$ by $-\beta$ in eqs. (1) and (2).
But we adopt a different approach. We set $\gamma=\alpha+\beta$ so that $\beta=\gamma-\alpha.$
Then eqs. (1) and (2) reduce to

 $\sin{(\gamma)}=\sin{\alpha}\cos{(\gamma-\alpha)}+\cos{\alpha}\sin{(\gamma-% \alpha)}$ (14)

and

 $\cos{\gamma}=\cos{\alpha}\cos{(\gamma-\alpha)}-\sin{\alpha}\sin{(\gamma-\alpha% )},$ (15)

respectively. We obtain the trigonometric difference formulae by solving eqs. (14) and (15) simultaneously.
$(\ref{eq:trisinsum1})\times\cos{\alpha}-(\ref{eq:tricossum1})\times\sin{\alpha}$ yields

 $\sin{\gamma}\cos{\alpha}-\cos{\gamma}\sin{\alpha}=(\sin^{2}{\alpha}+\cos^{2}{% \alpha})\ \sin{(\gamma-\alpha)}$

so that

 $\sin{(\gamma-\alpha)}=\sin{\gamma}\cos{\alpha}-\cos{\gamma}\sin{\alpha}$ (16)

using eq. (5).
Similarly, $(\ref{eq:trisinsum1})\times\sin{\alpha}+(\ref{eq:tricossum1})\times\cos{\alpha}$ results in

 $\cos{(\gamma-\alpha)}=\cos{\gamma}\cos{\alpha}+\sin{\gamma}\sin{\alpha}.$ (17)

## References

Title Deriving the trigonometric addition formulae using area and cosine rule DerivingTheTrigonometricAdditionFormulaeUsingAreaAndCosineRule1 2013-03-11 19:55:01 2013-03-11 19:55:01 dkrbabajee (19083) (0) 1 dkrbabajee (0) Definition