Addition Formulae and Double Angle Formulae

6.5   Addition Formulae and Double Angle Formulae
  • The addition formulae that are used to find trigonometry ratios of addition angles are as follows:


\(\text{sin}(A+B)= \text{sin} \ A \text{ cos} \ B + \text{cos} \ A \text{ sin} \ B\)

\(\text{sin}(A-B)= \text{sin} \ A \text{ cos} \ B - \text{cos} \ A \text{ sin} \ B\)


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

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


\(\text{tan}(A+B) = \dfrac{\text{tan }A + \text{tan }B}{1-\text{tan }A \text{ tan }B}\)

\(\text{tan}(AB) = \dfrac{\text{tan }A - \text{tan }B}{1+\text{tan }A \text{ tan }B}\)

  • Angles in the form \((A+B) \text{ or }(A-B)\) are called addition angles
  • Angles in the form \(2A, \ 3A, \ ...\) are known as double angles
  • The same method can be used to verify the other addition formulae

Prove the identities for 

\(\text{sin }(90^{\circ}+A) = \text{cos } A\).


Without using a calculator, find the value of \(\text{tan }15^{\circ}\).

(a) \(\text{sin }(90^{\circ}+A) \\ = \text{sin }90^{\circ} \text{ cos }A + \text{ cos }90^{\circ} \text{ sin }A\\ =(1) \text{ cos }A + (0) \text{ sin }A\\ =\text{ cos }A \)

Recall that:

Angles sin cos tan
\(45^{\circ}\) \(\dfrac{1}{\sqrt2}\) \(\dfrac{1}{\sqrt2}\) \(1\)
\(60^{\circ}\) \(\dfrac{\sqrt3}{2}\) \(\dfrac{1}{2}\) \(\sqrt3\)
  \(\begin{aligned} &\text{tan }15^{\circ}\\ &=\text{tan }(60^{\circ} -45^{\circ})\\\\ &=\dfrac{\text{tan }60^{\circ} - \text{tan }45^{\circ}}{1+\text{tan }60^{\circ} \ \text{tan }45^{\circ}}\\\\ &=\dfrac{\sqrt 3 - 1}{1+ (\sqrt3)(1)}\\\\ &=\dfrac{\sqrt3 -1}{\sqrt 3+1}\\\\ &=2-\sqrt3 \end{aligned} \)
  • The double-angle formulae:
    \(\text{sin } 2A = 2 \text{ sin }A\text{ cos }A\)    

\(\text{cos } 2A = \text{ cos}^2 \ A - \text{ sin}^2 \ A\)

\(\text{cos } 2A = 2\text{ cos}^2 \ A - 1\)

\(\text{cos } 2A = 1- 2\text{ sin}^2 \ A \)

  \(\text{tan }2A = \dfrac{2\text{ tan }A}{1-\text{tan}^2 \ A}\)  
  • Other formulae involving double angles can be derived by induction

  • This relation can be used to prove half-angle formulae where \(\text{sin }\dfrac{A}{2}\) , \(\text{kos }\dfrac{A}{2}\) and \(\text{tan }\dfrac{A}{2}\) are expressed in terms of  \(\text{sin }A\) dan \(\text{cos }A\) as below:

    \(\text{sin }\dfrac{A}{2} = \pm \sqrt{\dfrac{1-\text{ cos }A}{2}}\)    
  \(\text{cos }\dfrac{A}{2} = \pm \sqrt{\dfrac{1+\text{ cos }A}{2}}\)  
  \(\text{tan }\dfrac{A}{2} = \pm \sqrt{\dfrac{\text{sin }A}{1+\text{ cos }A}}\)  
(a) Find the value for the expression \(2\text{ sin }15^{\circ} \text{ cos }15^{\circ}\) using the double-angle formulae.
(b) Prove that \(\text{tan }\dfrac{x}{2} = \dfrac{1-\text{ cos }x}{\text{sin }x}\).
(a) \(2\text{ sin }15^{\circ} \text{ cos }15^{\circ}\\ = \text{sin }2(15^{\circ})\\ =\text{sin }30^{\circ}\\ =\dfrac{1}{2}\)

\(\begin{aligned} \text{Sebelah kanan } &= \dfrac{1-\text{ cos }x}{\text{sin }x}\\\\ &= \dfrac{1-\begin{pmatrix} 1-2\text{ sin}^2 \ \dfrac{x}{2} \end{pmatrix}}{2 \text{ sin }\dfrac{x}{2}\text{ cos }\dfrac{x}{2}}\\\\ &= \dfrac{2\text{ sin}^2 \ \dfrac{x}{2} }{2 \text{ sin }\dfrac{x}{2}\text{ cos }\dfrac{x}{2}}\\\\ &=\dfrac{\text{sin }\dfrac{x}{2}}{\text{cos }\dfrac{x}{2}}\\\\ &=\text{tan }\dfrac{x}{2} \end{aligned}\)


Hence, it is proven that \(\text{tan }\dfrac{x}{2} = \dfrac{1-\text{ cos }x}{\text{sin }x}\).