Proof 1
\( \sin (\alpha-\beta) = \sin \alpha \cos \beta-\cos \alpha \sin \beta \)
\( \require{AMSsymbols} \begin{align} \angle RPN &= 90^{\circ}-\angle PNR \\ &= \angle RNB \\ &= \angle QON \\ &= \alpha \\ \sin(\alpha-\beta) &= \sin \angle MOP \\ &= \displaystyle \frac{MP}{OP} \\ &= \frac{MR-PR}{OP} \\ &= \frac{QN}{OP}-\frac{PR}{OP} \\ &= \frac{QN}{\color{red}{ON}} \times \frac{\color{red}{ON}}{OP}-\frac{PR}{\color{red}{PN}} \times \frac{\color{red}{PN}}{OP} \\ &= \sin \alpha \cos \beta-\cos \angle RPN \sin \beta \\ \therefore \sin (\alpha-\beta) &= \sin \alpha \cos \beta-\cos \alpha \sin \beta \end{align} \)
Proof 2
\( \cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \)
\( \require{AMSsymbols} \begin{align} \cos (\alpha-\beta) &= \displaystyle \frac{OM}{OP} \\ &= \frac{OQ+MQ}{OP} \\ &= \frac{OQ}{OP} + \frac{RN}{OP} \\ &= \frac{OQ}{\color{red}{ON}} \times \frac{\color{red}{ON}}{OP} + \frac{RN}{\color{red}{NP}} \times \frac{\color{red}{NP}}{OP} \\ &= \cos \alpha \cos \beta + \sin \angle RPN \sin \beta \\ \therefore \cos (\alpha-\beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{align} \)
Proof 3
\( \tan (\alpha-\beta) = \displaystyle \frac{\tan \alpha-\tan \beta}{1 + \tan \alpha \tan \beta} \)
\( \require{AMSsymbols} \begin{align} \tan (\alpha-\beta) &= \displaystyle \frac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)} \\ &= \frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} \\ &= \frac{\displaystyle \frac{\sin \alpha \cos \beta}{\color{red}{\cos \alpha \cos \beta}}-\frac{\displaystyle \cos \alpha \sin \beta}{\color{red}{\cos \alpha \cos \beta}}}{\displaystyle \frac{\cos \alpha \cos \beta}{\color{red}{\cos \alpha \cos \beta}} + \frac{\displaystyle \sin \alpha \sin \beta}{\color{red}{\cos \alpha \cos \beta}}} \\ &= \frac{\displaystyle \frac{\sin \alpha}{\cos \alpha}-\frac{\sin \beta}{\cos \beta}}{1+ \displaystyle \frac{\sin \alpha}{\cos \alpha} \times \frac{\sin \beta}{\cos \beta}} \\ &= \displaystyle \frac{\tan \alpha-\tan \beta}{1 + \tan \alpha \tan \beta} \\ \therefore \tan (\alpha-\beta) &= \displaystyle \frac{\tan \alpha-\tan \beta}{1 + \tan \alpha \tan \beta} \end{align} \)
Example 1
Find the exact value of \( \sin 15^{\circ} \).
\( \require{AMSsymbols} \begin{align} \sin 15 ^{\circ} &= \sin (45^{\circ}-30^{\circ}) \\ &= \sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\ &= \displaystyle \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ &= \frac{\sqrt{3}-1}{2 \sqrt{2}} \end{align} \)
Example 2
Find the exact value of \( \cos 15^{\circ} \).
\( \require{AMSsymbols} \begin{align} \cos 15 ^{\circ} &= \cos (45^{\circ}-30^{\circ}) \\ &= \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} \\ &= \displaystyle \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \\ &= \frac{\sqrt{3}+1}{2 \sqrt{2}} \end{align} \)
Example 3
Find the exact value of \( \tan 15^{\circ} \).
\( \require{AMSsymbols} \begin{align} \tan 15 ^{\circ} &= \tan (45^{\circ}-30^{\circ}) \\ &= \displaystyle \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1 + \tan 45^{\circ} \times \tan 30 ^{\circ}} \\ &= \frac{1-\displaystyle \frac{1}{\sqrt{3}}}{1 + \displaystyle 1 \times \frac{1}{\sqrt{3}}} \\ &= \frac{\sqrt{3}-1}{\sqrt{3}+1} \\ &= \frac{(\sqrt{3}-1)^2}{3-1} \\ &= \frac{4-2\sqrt{3}}{2} \\ &= 2-\sqrt{3} \end{align} \)
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