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rosijanka [135]
3 years ago
8

How do you solve cos(pi/24) using Half-Angle formulas, and leaving in simplified form?

Mathematics
1 answer:
adoni [48]3 years ago
6 0
\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}
\\
\\ \cos{\frac{\pi}{24}}=\cos{\frac{\frac{\pi}{12}}{2}}=\sqrt{\frac{1+\cos \frac{\pi}{12}}{2}}= \sqrt{ \frac{1}{2}+ \frac{\cos \frac{\pi}{12}}{2}  } 
\\
\\ \cos{\frac{\pi}{12}}=\cos{\frac{\frac{\pi}{6}}{2}}=\sqrt{\frac{1+\cos \frac{\pi}{6}}{2}}= \sqrt{ \frac{1}{2}+ \frac{ \frac{ \sqrt{3} }{2} }{2}  } =\sqrt{ \frac{2}{4}+  \frac{ \sqrt{3} }{4}   } = \sqrt{\frac{ 2+\sqrt{3} }{4} } = 
\\
\\ =\frac{ \sqrt{2+\sqrt{3}} }{2} 


\cos{\frac{\pi}{24}}= \sqrt{ \frac{1}{2}+ \frac{\cos \frac{\pi}{12}}{2}  } = \sqrt{ \frac{1}{2}+ \frac{\frac{ \sqrt{2+\sqrt{3}} }{2}}{2}  } = \sqrt{ \frac{2}{4}+ \frac{ \sqrt{2+\sqrt{3}} }{4}}  } =\sqrt{ \frac{ 2+\sqrt{2+\sqrt{3}} }{4}}  } 
\\
\\\cos{\frac{\pi}{24}}=\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}} {2}


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