Solveeit Logo

Question

Question: If \(\alpha = 22{^\circ}30',\) then \((1 + \cos\alpha)(1 + \cos 3\alpha)\) \((1 + \cos 5\alpha)(1 + ...

If α=2230,\alpha = 22{^\circ}30', then (1+cosα)(1+cos3α)(1 + \cos\alpha)(1 + \cos 3\alpha) (1+cos5α)(1+cos7α)(1 + \cos 5\alpha)(1 + \cos 7\alpha) equals

A

1/8

B

1/4

C

1+222\frac{1 + \sqrt{2}}{2\sqrt{2}}

D

212+1\frac{\sqrt{2} - 1}{\sqrt{2} + 1}

Answer

1/8

Explanation

Solution

We know, sin221o2=1222\sin 22\frac{1^{o}}{2} = \frac{1}{2}\sqrt{2 - \sqrt{2}}and

cos221o2=122+2\cos 22\frac{1^{o}}{2} = \frac{1}{2}\sqrt{2 + \sqrt{2}}

(1+cos221o2)(1+cos671o2)(1+cos1121o2)\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)

(1+cos1571o2)\left( 1 + \cos 157\frac{1^{o}}{2} \right)

=(1+122+2)(1+1222)(11222)= \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)

(1122+2)\left( 1 - \frac{1}{2}\sqrt{2 + \sqrt{2}} \right)

=[114(2+2)][114(22)]= \left\lbrack 1 - \frac{1}{4}(2 + \sqrt{2}) \right\rbrack\left\lbrack 1 - \frac{1}{4}(2 - \sqrt{2}) \right\rbrack =(422)(42+2)16= \frac{(4 - 2 - \sqrt{2})(4 - 2 + \sqrt{2})}{16}

=(22)(2+2)16=4216=18= \frac{(2 - \sqrt{2})(2 + \sqrt{2})}{16} = \frac{4 - 2}{16} = \frac{1}{8}.