Notes: Find the exact value of cos pi/16

Find the exact value of $\displaystyle\frac{{\cos{\pi}}}{{16}}$
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We use a half-angle formula: $\displaystyle{\cos{{\left(\frac{\alpha}{{2}}\right)}}}=\sqrt{{\frac{{1}}{{2}}{\left({1}+{\cos{}}\right.}}}$
alpha))

(1) $\displaystyle\frac{{\cos{\pi}}}{{8}}=\frac{{\cos{{\left(\frac{\pi}{{4}}\right)}}}}{{2}}=\sqrt{{\frac{{1}}{{2}}{\left({1}+\frac{{\cos{\pi}}}{{4}}\right)}}}$

$\displaystyle=\sqrt{{\frac{{1}}{{2}}{\left({1}+\frac{\sqrt{{{2}}}}{{2}}\right)}}}$

$\displaystyle=\sqrt{{\frac{{{2}+\sqrt{{{2}}}}}{{4}}}}$

$\displaystyle=\frac{\sqrt{{{2}+\sqrt{{{2}}}}}}{{2}}$

(2) $\displaystyle\frac{{\cos{\pi}}}{{16}}=\frac{{\cos{{\left(\frac{\pi}{{8}}\right)}}}}{{2}}$

$\displaystyle=\sqrt{{\frac{{1}}{{2}}{\left({1}+\frac{{\cos{\pi}}}{{8}}\right)}}}$

$\displaystyle=\sqrt{{\frac{{1}}{{2}}{\left(\frac{{\sqrt{{{2}+\sqrt{{{2}}}}}}}{{2}}\right)}}}$

$\displaystyle=\sqrt{{\frac{{{2}+\sqrt{{{2}+\sqrt{{{2}}}}}}}{{4}}}}$

$\displaystyle=\frac{{\sqrt{{{2}+\sqrt{{{2}+\sqrt{{{2}}}}}}}}}{{2}}$

Thus
$\displaystyle\frac{{\cos{\pi}}}{{16}}=\frac{{\sqrt{{{2}+\sqrt{{{2}+\sqrt{{{2}}}}}}}}}{{2}}\approx{.9807852804}$

Notes

Monday, 4 September 2017

Find the exact value of cos pi/16

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