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Question

Mathematics Question on Trigonometric Functions

What is the value of (sin2212+cos2212)4\left( \sin 22 \frac{1^{\circ}}{2} + \cos \, 22 \frac{1^{\circ} }{2} \right)^4 ?

A

3+222\frac{3 + 2\sqrt{2}}{2}

B

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

C

32+22\frac{3\sqrt{2} + 2 }{2}

D

1

Answer

3+222\frac{3 + 2\sqrt{2}}{2}

Explanation

Solution

Let x=(sin2212+cos22212)4x = \left(\sin22 \frac{1^{\circ}}{2} + \cos^{2} 22 \frac{1^{\circ}}{2}\right)^{4} = \left\\{\left(\sin22 \frac{1^{\circ}}{2} + \cos22 \frac{1}{2}^{\circ}\right)^{2}\right\\}^{2} =(sin22212+cos222122sin2212cos2212)2= \left(\sin^{2} 22 \frac{1^{\circ}}{2} + \cos^{2} 22 \frac{1^{\circ}}{2} 2 \sin 22 \frac{1^{\circ}}{2} \cos22 \frac{1^{\circ}}{2}\right)^{2} =(1+sin45)2= \left(1+\sin45^{\circ}\right)^{2} =(1+12)2=(2+12)2= \left(1 + \frac{1}{\sqrt{2}}\right)^{2} = \left( \frac{\sqrt{2} + 1}{\sqrt{2}}\right)^{2} =2+1+222=3+222= \frac{2+1+2\sqrt{2}}{2} = \frac{3+2\sqrt{2}}{2}