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Question

Mathematics Question on Trigonometric Functions

The value of cos⁡221∘2\cos 22 \frac{1^{\circ}}{2} is

A

2+12\sqrt{\frac{\sqrt{2}+ 1}{2}}

B

2+12\sqrt{\frac{\sqrt{2}+ 1}{\sqrt{2}}}

C

2+122\sqrt{\frac{\sqrt{2}+ 1}{2 \sqrt{2}}}

D

2+122\frac{\sqrt{2}+ 1}{2 \sqrt{2}}

Answer

2+122\sqrt{\frac{\sqrt{2}+ 1}{2 \sqrt{2}}}

Explanation

Solution

We have, cos⁡A2=±1+cos⁡A2\cos \frac{A}{2} = \pm \sqrt{ \frac{1 + \cos A}{2}} Putting A=45∘A = 45^{\circ} , we get cos⁡221∘2=1+cos⁡45∘2\cos 22 \frac{1^{\circ}}{2} = \sqrt{\frac{1 + \cos 45^{\circ}}{2}} [∵cos⁡221∘2is+ve]\left[ \because \cos22 \frac{1^{\circ}}{2} is + ve \right] ⇒cos⁡221∘2=1+1/22=2+122\Rightarrow \cos22 \frac{1^{\circ}}{2} = \sqrt{\frac{1+1/\sqrt{2}}{2}} = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}} }