Expert Answer to "How can I find the value ofsin24π24+cos24π24"

Beryneingmk39

Answered question

2022-03-24

How can I find the value of

\sin^{24} \frac{π}{24} + \cos^{24} \frac{π}{24}

Mason Knight

Beginner2022-03-25Added 11 answers

I would use complex exponentials. We have:

\sin^{24} \frac{π}{24} + \cos^{24} \frac{π}{24} = {(\frac{e^{i \frac{π}{24}} - e^{- i \frac{π}{24}}}{2 i})}^{24} + {(\frac{e^{i \frac{π}{24}} + e^{- i \frac{π}{24}}}{2})}^{24})^{24}

= \frac{1}{2^{24}} (\sum_{k = 0}^{24} (\begin{matrix} 24 \\ k \end{matrix}) {(- 1)}^{k} e^{i \frac{24 - 2 k}{24} π}) + \frac{1}{2^{24} (\sum_{k = 0}^{24} (\begin{matrix} 24 \\ k \end{matrix}) e^{i \frac{24 - 2 k}{24} π})}

= \frac{1}{2^{23}} \sum_{l = 0}^{12} (\begin{matrix} 24 \\ 2 l \end{matrix}) e^{i \frac{12 - 2 l}{12} π}

after canceling every other term in the two sums. We can rewrite this last expression in terms of trig functions again, as

\frac{1}{2^{22}} (\sum_{m = 0}^{5} (\begin{matrix} 24 \\ 2 m \end{matrix}) \cos (π - \frac{m π}{6}) + \frac{1}{2} (\begin{matrix} 24 \\ 12 \end{matrix}))

which simplifies into

\frac{1}{2^{22}} (\frac{3428999}{2} + 1960980 \frac{\sqrt{3}}{2})

= \frac{3428999 + 1960980 \sqrt{3}}{2^{23}}

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