Addition Formulae and Double Angle Formulae

6.5

Addition Formulae and Double Angle Formulae

The image is a diagram titled 'Form of Angles' by Pandai. It is divided into two categories: 'Addition Angles' and 'Double Angles.' The 'Addition Angles' section includes formulas (A+B) or (A-B), while the 'Double Angles' section includes multiples like 2A, 3A, etc. Each category is enclosed in a separate box, with arrows pointing to them from the central title.

Addition Formulae

\(\sin{(A\pm B)}=\sin{A}\cos{B}\pm \cos{A}\sin{B}\)

\(\cos{(A\pm B)}=\cos{A}\cos{B}\mp \sin{A}\sin{B}\)

\(\tan{(A\pm B)}=\dfrac{\tan{A}\pm \tan{B}}{1\mp \tan{A}\tan{B}}\)

Double Angle Formulae

\(\sin{2A}=2\sin{A}\cos{A}\)

\(\begin{aligned} \cos{2A}&= \cos^2{A}-\sin^2{A} \\ &=2\cos^2{A}-1 \\ &=1-2\sin^2{A} \end{aligned}\)

\(\tan{2A}=\dfrac{2\tan{A}}{1-\tan^2{A}}\)

Half Angle Formulae

\(\sin{\dfrac{A}{2}}=\pm\sqrt{\dfrac{1-\cos{A}}{2}}\)

\(\cos{\dfrac{A}{2}}=\pm\sqrt{\dfrac{1+\cos{A}}{2}}\)

\(\tan{\dfrac{A}{2}}=\pm\sqrt{\dfrac{\sin{A}}{1+\cos{A}}}\)

Example \(1\)

Question

(a)	Prove the identities for
	\(\text{sin }(90^{\circ}+A) = \text{cos } A\)

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

Solution

(a)

\(\begin{aligned} \text{sin }(90^{\circ}+A) &= \text{sin }90^{\circ} \text{ cos }A + \text{ cos }90^{\circ} \text{ sin }A\\ &=(1) \text{ cos }A + (0) \text{ sin }A\\ &=\text{ cos }A. \end{aligned}\)

(b)

Recall that:

Angles	sin	cos	tan
\(45^\circ\)	\(\dfrac{1}{\sqrt{2}}\)	\(\dfrac{1}{\sqrt{2}}\)	\(1\)
\(60^\circ\)	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{1}{2}\)	\(\sqrt{3}\)

\(\begin{aligned} \text{tan }15^{\circ}&= \tan{(60^\circ-45^\circ)} \\\\ \text{tan }(60^{\circ} -45^{\circ})&=\dfrac{\text{tan }60^{\circ} - \text{tan }45^{\circ}}{1+\text{tan }60^{\circ} \ \text{tan }45^{\circ}}\\\\ &=\dfrac{\sqrt 3 - 1}{1+ (\sqrt3)(1)}\\\\ &=\dfrac{\sqrt3 -1}{\sqrt 3+1}\\\\ &=2-\sqrt3. \end{aligned}\)

Example \(2\)

Question

(a)	Find the value for the expression \(2\sin{15^\circ}\cos{15^\circ}\) using the double-angle formulae.

(b)	Prove that \(\text{tan }\dfrac{x}{2} = \dfrac{1-\text{ cos }x}{\text{sin }x}\).

Solution

(a)

\(\begin{aligned} 2\sin{15^\circ}\cos{15^\circ}&=\sin{2(15^\circ)} \\ &=\sin{30^\circ} \\ &=\dfrac{1}{2}. \end{aligned}\)

(b)

From right-hand side:

\(\begin{aligned} \dfrac{1-\cos{x}}{\sin{x}}&=\dfrac{1-\left( 1-2\sin^2{\dfrac{x}{2}} \right)}{2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}} \\\\ &=\dfrac{2\sin^2{\dfrac{x}{2}}}{2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}} \\\\ &=\dfrac{\sin{\dfrac{x}{2}}}{\cos{\dfrac{x}{2}}} \\\\ &=\tan{\dfrac{x}{2}}. \end{aligned}\)

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