Question #139926

Marion needs to find the cosine of Pi over 12. If she knows that cos Pi over 6 = root 3 over 2, how can she use this fact to find the cosine of Pi over 12? What is her answer. Please explain and show your work

Expert's answer

Use double angle formula for cosine cos(2A)=2cos2(A)1cos(2A)=2cos^2(A)-1


So, the cosine of angle AA can be written as,


2cos2(A)=1+cos(2A)2cos^2(A)=1+cos(2A)


cos2(A)=12(1+cos(2A))cos^2(A)=\frac{1}{2}(1+cos(2A))


cos(A)=12(1+cos(2A))cos(A)=\sqrt{\frac{1}{2}(1+cos(2A))}


Plug A=π12A=\frac{\pi}{12} into the relation cos(A)=12(1+cos(2A))cos(A)=\sqrt{\frac{1}{2}(1+cos(2A))} to obtain,


cos(π12)=12(1+cos(2(π12)))cos(\frac{\pi}{12})=\sqrt{\frac{1}{2}(1+cos(2(\frac{\pi}{12})))}


=12(1+cos(π6))=\sqrt{\frac{1}{2}(1+cos(\frac{\pi}{6}))}


=12(1+32)=\sqrt{\frac{1}{2}(1+\frac{\sqrt{3}}{2})} .....plug cos(π6)=32cos(\frac{\pi}{6})=\frac{\sqrt3}{2}


=12(2+32)=\sqrt{\frac{1}{2}(\frac{2+\sqrt{3}}{2})}


=14(2+3)=\sqrt{\frac{1}{4}(2+\sqrt{3})}


=122+30.965926=\frac{1}{2}\sqrt{2+\sqrt{3}}\approx0.965926


Therefore, the cosine of π12\frac{\pi}{12} is cos(π12)=122+30.965926cos(\frac{\pi}{12})=\frac{1}{2}\sqrt{2+\sqrt{3}}\approx0.965926

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