જો alpha = 22^circ 30' તો (1 + cos alpha | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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}$.

જો $\alpha = 22^\circ 30' $ તો $(1 + \cos \alpha )(1 + \cos 3\alpha )$ $(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ = . . .. .

Similar Questions

જો $\sin \theta = - \frac{1}{{\sqrt 2 }}$ અને $\tan \theta = 1,$ તો $\theta $ એ ક્યાં ચરણમાં છે ?

આપેલ પૈકી ક્યો સંબધ શક્ય છે ?

$\cos 15^\circ = $

સાબિત કરો કે : $\sin x+\sin 3 x+\sin 5 x+\sin 7 x=4 \cos x \cos 2 x \sin 4 x$

$\sin \frac{31 \pi}{3}$ નું મૂલ્ય શોધો.