How do you find exactly ${sin 78.75}^{o}$ $sin 78.75^o$ , in a closed form?

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Answer 1 · 2016-11-09T02:27:00

$\sqrt{⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1 + \sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}}}{2} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠}$ $sqrt(((1+sqrt((1+1/sqrt 2)/2))/2)$ =0.98078528, nearly.

$sin ({78.75}^{o})$ $sin (78.75^o)$

$= cos (90^{o} - {78.75}^{o})$ $=cos(90^o-78.75^o)$

$= cos ({11.25}^{o})$ $=cos (11.25^o)$

$= \sqrt{(\frac{1 + cos ({22.5}^{o})}{2})}$ $=sqrt(((1+cos(22.5^o))/2)$

$= \sqrt{⎛ ⎜ ⎜ ⎝ \frac{1 + \sqrt{\frac{1 + cos (45^{o})}{2}}}{2} ⎞ ⎟ ⎟ ⎠}$ $=sqrt(((1+sqrt((1+cos(45^o)}/2))/2)$

Note for information:: A completed joint paper by me and two

others in USA, presenting an innovative generalizaion of this

method, is on the rounds, expecting publication in a repute journal.

How do you find exactly sin 78.75^osin78.75osin 78.75^o, in a closed form?