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Question Number 133587 by bemath last updated on 23/Feb/21

$$\mathrm{Find}\mathrm{the}\mathrm{exact}\mathrm{value}\mathrm{of}\cos 12°.$$

Answered by Dwaipayan Shikari last updated on 23/Feb/21

$$sin\frac{\pi }{60}=sin\frac{\pi }{10}cos\frac{\pi }{12}-sin\frac{\pi }{12}cos\frac{\pi }{10}$$$$=\frac{\sqrt{5}-1}{4}.\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{\sqrt{10-2\sqrt{5}}}{4}.\frac{\sqrt{2+\sqrt{3}}}{2}$$$$=\frac{1}{16}((\sqrt{5}-1)(\sqrt{6}-\sqrt{2})-(\sqrt{6}+\sqrt{2})\sqrt{10-2\sqrt{5}})$$$$cos\frac{\pi }{60}=\sqrt{1-{sin}^2\frac{\pi }{60}}$$$$cos\frac{\pi }{15}=cos\frac{\pi }{12}cos\frac{\pi }{60}+sin\frac{\pi }{60}sin\frac{\pi }{12}$$$$=\frac{\sqrt{6}+\sqrt{2}}{2}cos\frac{\pi }{60}+\frac{\sqrt{6}-\sqrt{2}}{2}sin\frac{\pi }{60}$$$$=\frac{1}{8}(\sqrt{5}-1+\sqrt{6(5+\sqrt{5})})$$

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