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If $\alpha = 22^\circ 30',$ then $(1 + \cos \alpha )(1 + \cos 3\alpha )$ $(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ equals
$1/8$
$1/4$
$\frac{{1 + \sqrt 2 }}{{2\sqrt 2 }}$
$\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}$
Solution
(a) We know, $\sin 22\frac{{{1^o}}}{2} = \frac{1}{2}\sqrt {2 – \sqrt 2 } $
and $\cos 22\frac{{{1^o}}}{2} = \frac{1}{2}\sqrt {2 + \sqrt 2 } $
$\therefore \left( {1 + \cos 22\frac{{{1^o}}}{2}} \right)\,\left( {1 + \cos 67\frac{{{1^o}}}{2}} \right)\,\left( {1 + \cos 112\frac{{{1^o}}}{2}} \right)$
$\left( {1 + \cos 157\frac{{{1^o}}}{2}} \right)$
$ = \left( {1 + \frac{1}{2}\sqrt {2 + \sqrt 2 } } \right)\,\left( {1 + \frac{1}{2}\sqrt {2 – \sqrt 2 } } \right)\,\left( {1 – \frac{1}{2}\sqrt {2 – \sqrt 2 } } \right)\,$
$\left( {1 – \frac{1}{2}\sqrt {2 + \sqrt 2 } } \right)$
$ = \left[ {1 – \frac{1}{4}(2 + \sqrt 2 )} \right]\,\left[ {1 – \frac{1}{4}(2 – \sqrt 2 )} \right]$
$ = \frac{{(4 – 2 – \sqrt 2 )(4 – 2 + \sqrt 2 )}}{{16}}$
$ = \frac{{(2 – \sqrt 2 )(2 + \sqrt 2 )}}{{16}} = \frac{{4 – 2}}{{16}} = \frac{1}{8}$.
