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

Find the exact value of cos 12° .

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

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

sin(π/(60))=sin(π/(10))cos(π/(12))−sin(π/(12))cos(π/(10))  =(((√5)−1)/4).((√(2+(√3)))/2)−((√(10−2(√5)))/4).((√(2+(√3)))/2)  =(1/(16))(((√5)−1)((√6)−(√2))−((√6)+(√2))(√(10−2(√5))))  cos(π/(60))=(√(1−sin^2 (π/(60))))  cos(π/(15))=cos(π/(12))cos(π/(60))+sin(π/(60))sin(π/(12))  =(((√6)+(√2))/2)cos(π/(60))+(((√6)−(√2))/2)sin(π/(60))  =(1/8)((√5)−1+(√(6(5+(√5)))))

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

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