## 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

Example

(a)

Prove the identities for

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

(b)

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

Solution:

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

(b)

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:

 $$\text{sin }\dfrac{A}{2} = \pm \sqrt{\dfrac{1-\text{ cos }A}{2}}$$ $$\text{cos }\dfrac{A}{2} = \pm \sqrt{\dfrac{1+\text{ cos }A}{2}}$$ $$\text{tan }\dfrac{A}{2} = \pm \sqrt{\dfrac{\text{sin }A}{1+\text{ cos }A}}$$

Example

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