\(\sin{(A\pm B)}=\sin{A}\,\text{kos}\,B \pm \text{kos}\,A\sin{B}\)

\(\text{kos}\,(A\pm B)=\text{kos}\,A\,\text{kos}\,B\mp \sin{A}\sin{B}\)

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

\(\sin{2A}=2\sin{A}\,\text{kos}\,A\)

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

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

\(\sin{\dfrac{A}{2}}=\pm\sqrt{\dfrac{1-\text{kos}\,A}{2}}\)

\(\text{kos}\,\dfrac{A}{2}=\pm\sqrt{\dfrac{1+\text{kos}\,A}{2}}\)

\(\tan{\dfrac{A}{2}}=\pm\sqrt{\dfrac{\sin{A}}{1+\text{kos}\,A}}\)

(a)

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

(b)

Ingat kembali bahawa:

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

(a)

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

(b)

Daripada sebelah kanan:

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

Maka, terbukti bahawa \(\text{tan }\dfrac{x}{2} = \dfrac{1-\text{ kos }x}{\text{sin }x}\).

\(\sin{(A\pm B)}=\sin{A}\,\text{kos}\,B \pm \text{kos}\,A\sin{B}\)

\(\text{kos}\,(A\pm B)=\text{kos}\,A\,\text{kos}\,B\mp \sin{A}\sin{B}\)

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

\(\sin{2A}=2\sin{A}\,\text{kos}\,A\)

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

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

\(\sin{\dfrac{A}{2}}=\pm\sqrt{\dfrac{1-\text{kos}\,A}{2}}\)

\(\text{kos}\,\dfrac{A}{2}=\pm\sqrt{\dfrac{1+\text{kos}\,A}{2}}\)

\(\tan{\dfrac{A}{2}}=\pm\sqrt{\dfrac{\sin{A}}{1+\text{kos}\,A}}\)

(a)

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

(b)

Ingat kembali bahawa:

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

(a)

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

(b)

Daripada sebelah kanan:

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

Maka, terbukti bahawa \(\text{tan }\dfrac{x}{2} = \dfrac{1-\text{ kos }x}{\text{sin }x}\).