Question

match each trigonometric expression with its equivalent form. each may only be used exactly once.

1. \\( \cos ( \frac { 13 \pi } { 12 } ) \\)
2. \\( \tan ( \frac { 11 \pi } { 12 } ) \\)
3. \\( \cos ( \frac { \pi } { 9 } ) \sin ( \frac { 7 \pi } { 18 } ) + \sin ( \frac { \pi } { 9 } ) \cos ( \frac { 7 \pi } { 18 } ) \\)
4. \\( \cos ( \frac { 4 \pi } { 9 } ) \cos ( \frac { \pi } { 18 } ) - \sin ( \frac { \pi } { 18 } ) \sin ( \frac { 4 \pi } { 9 } ) \\)
5. \\( 2 \sin ( \frac { \pi } { 12 } ) \cos ( \frac { \pi } { 12 } ) \\)

Explanation:

Step1: Use trigonometric identities

• For \(2\sin(\frac{\pi}{12})\cos(\frac{\pi}{12})\), use the double - angle formula \(\sin(2\alpha)=2\sin\alpha\cos\alpha\). Here \(\alpha = \frac{\pi}{12}\), so \(2\sin(\frac{\pi}{12})\cos(\frac{\pi}{12})=\sin(2\times\frac{\pi}{12})=\sin(\frac{\pi}{6})=\frac{1}{2}\). So \(\frac{1}{2}\) matches with \(5.2\sin(\frac{\pi}{12})\cos(\frac{\pi}{12})\).
• For \(\cos(\frac{\pi}{9})\sin(\frac{7\pi}{18})+\sin(\frac{\pi}{9})\cos(\frac{7\pi}{18})\), use the sum formula \(\sin(A + B)=\sin A\cos B+\cos A\sin B\). Let \(A=\frac{\pi}{9}\) and \(B = \frac{7\pi}{18}\), then \(A + B=\frac{\pi}{9}+\frac{7\pi}{18}=\frac{2\pi + 7\pi}{18}=\frac{\pi}{2}\). So \(\cos(\frac{\pi}{9})\sin(\frac{7\pi}{18})+\sin(\frac{\pi}{9})\cos(\frac{7\pi}{18})=\sin(\frac{\pi}{2})\). So \(\sin(\frac{\pi}{2})\) matches with \(3.\cos(\frac{\pi}{9})\sin(\frac{7\pi}{18})+\sin(\frac{\pi}{9})\cos(\frac{7\pi}{18})\).
• For \(\cos(\frac{4\pi}{9})\cos(\frac{\pi}{18})-\sin(\frac{\pi}{18})\sin(\frac{4\pi}{9})\), use the cosine of sum formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\). Let \(A=\frac{4\pi}{9}\) and \(B=\frac{\pi}{18}\), then \(A + B=\frac{8\pi+\pi}{18}=\frac{\pi}{2}\). So \(\cos(\frac{4\pi}{9})\cos(\frac{\pi}{18})-\sin(\frac{\pi}{18})\sin(\frac{4\pi}{9})=\cos(\frac{\pi}{2})\). So \(\cos(\frac{\pi}{2})\) matches with \(4.\cos(\frac{4\pi}{9})\cos(\frac{\pi}{18})-\sin(\frac{\pi}{18})\sin(\frac{4\pi}{9})\).
• For \(\tan(\frac{11\pi}{12})\), use the formula \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\). \(\tan(\frac{11\pi}{12})=\tan(\pi-\frac{\pi}{12})=-\tan(\frac{\pi}{12})\). And \(\tan(\frac{\pi}{12})=\tan(\frac{\pi}{3}-\frac{\pi}{4})=\frac{\tan\frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}}=\frac{\sqrt{3}-1}{1 + \sqrt{3}}=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3 - 2\sqrt{3}+1}{2}=2-\sqrt{3}\), so \(\tan(\frac{11\pi}{12})=\sqrt{3}-2\). So \(\sqrt{3}-2\) matches with \(2.\tan(\frac{11\pi}{12})\).
• For \(\cos(\frac{13\pi}{12})\), use the formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\). \(\cos(\frac{13\pi}{12})=\cos(\pi+\frac{\pi}{12})=-\cos(\frac{\pi}{12})\). And \(\cos(\frac{\pi}{12})=\frac{\sqrt{6}+\sqrt{2}}{4}\), so \(\cos(\frac{13\pi}{12})=-\frac{\sqrt{2}+\sqrt{6}}{4}\). So \(-\frac{\sqrt{2}+\sqrt{6}}{4}\) matches with \(1.\cos(\frac{13\pi}{12})\).

Answer:

• \(\cos(\frac{\pi}{2})\) matches with \(4.\cos(\frac{4\pi}{9})\cos(\frac{\pi}{18})-\sin(\frac{\pi}{18})\sin(\frac{4\pi}{9})\)
• \(\frac{1}{2}\) matches with \(5.2\sin(\frac{\pi}{12})\cos(\frac{\pi}{12})\)
• \(\sqrt{3}-2\) matches with \(2.\tan(\frac{11\pi}{12})\)
• \(\sin(\frac{\pi}{2})\) matches with \(3.\cos(\frac{\pi}{9})\sin(\frac{7\pi}{18})+\sin(\frac{\pi}{9})\cos(\frac{7\pi}{18})\)
• \(-\frac{\sqrt{2}+\sqrt{6}}{4}\) matches with \(1.\cos(\frac{13\pi}{12})\)