If alpha = 22^circ 30', then (1 + cos alp | vedclass

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Question

(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 } $

$

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(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 } $

$	herefore \left( {1 + \cos 22\frac{{{1^o}}}{2}} ight)\,\left( {1 + \cos 67\frac{{{1^o}}}{2}} ight)\,\left( {1 + \cos 112\frac{{{1^o}}}{2}} ight)$

$\left( {1 + \cos 157\frac{{{1^o}}}{2}} ight)$

$ = \left( {1 + \frac{1}{2}\sqrt {2 + \sqrt 2 } } ight)\,\left( {1 + \frac{1}{2}\sqrt {2 - \sqrt 2 } } ight)\,\left( {1 - \frac{1}{2}\sqrt {2 - \sqrt 2 } } ight)\,$

$\left( {1 - \frac{1}{2}\sqrt {2 + \sqrt 2 } } ight)$

$ = \left[ {1 - \frac{1}{4}(2 + \sqrt 2 )} ight]\,\left[ {1 - \frac{1}{4}(2 - \sqrt 2 )} ight]$

$ = \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}$.

If $\alpha = 22^\circ 30',$ then $(1 + \cos \alpha )(1 + \cos 3\alpha )$ $(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ equals

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