A) 1/8
B) 1/4
C) \[\frac{1+\sqrt{2}}{2\sqrt{2}}\]
D) \[\frac{\sqrt{2}-1}{\sqrt{2}+1}\]
Correct Answer: A
Solution :
We know, \[\sin 22\frac{{{1}^{o}}}{2}=\frac{1}{2}\sqrt{2-\sqrt{2}}\] and \[\cos 22\frac{{{1}^{o}}}{2}=\frac{1}{2}\sqrt{2+\sqrt{2}}\] \[\therefore \left( 1+\cos 22\frac{{{1}^{o}}}{2} \right)\,\left( 1+\cos 67\frac{{{1}^{o}}}{2} \right)\,\left( 1+\cos 112\frac{{{1}^{o}}}{2} \right)\]\[\left( 1+\cos 157\frac{{{1}^{o}}}{2} \right)\] \[=\left( 1+\frac{1}{2}\sqrt{2+\sqrt{2}} \right)\,\left( 1+\frac{1}{2}\sqrt{2-\sqrt{2}} \right)\,\left( 1-\frac{1}{2}\sqrt{2-\sqrt{2}} \right)\,\]\[\left( 1-\frac{1}{2}\sqrt{2+\sqrt{2}} \right)\] \[=\left[ 1-\frac{1}{4}(2+\sqrt{2}) \right]\,\left[ 1-\frac{1}{4}(2-\sqrt{2}) \right]\] \[=\frac{(4-2-\sqrt{2})(4-2+\sqrt{2})}{16}\] \[=\frac{(2-\sqrt{2})(2+\sqrt{2})}{16}=\frac{4-2}{16}=\frac{1}{8}\].
You need to login to perform this action.
You will be redirected in
3 sec
