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Geometry Level 3

cos ( 7. 5 ) = ? \large \cos(7.5^\circ) = \, ?

3 + 2 + 2 4 \frac{\sqrt{\sqrt{\sqrt{3} + 2} + 2}}{4} 3 2 \frac{\sqrt{3}}{2} 3 + 1 3 \frac{\sqrt{3} + 1}{3} 3 + 2 4 \frac{\sqrt{3} + \sqrt{2}}{4} 3 + 2 8 \frac{\sqrt{3} + 2}{8} 3 + 2 2 \frac{\sqrt{\sqrt{3} + 2}}{2} 3 + 2 + 2 2 \frac{\sqrt{\sqrt{\sqrt{3} + 2} + 2}}{2}

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1 solution

Milan Milanic
Jan 11, 2016

First of all, I must say that @Drex Beckman , with his solution for one of my problems forced me to think in this direction. Therefore, I think it is appropriate to put his name in here :)

Solution: There is a formula which states: c o s x = c o s 2 x + 1 2 cos x = \sqrt{ \frac{cos 2x + 1}{2} } . I think it is a common knowledge that c o s 30 ° = 3 2 cos 30° = \frac{\sqrt{3}}{2} . Apply this formula, to get c o s 15 ° cos 15° and afterwards to get c o s 7.5 ° cos 7.5° . The first one is 3 + 2 2 \frac{\sqrt{\sqrt{3} + 2}}{2} and the latter one is 3 + 2 + 2 2 \frac{\sqrt{\sqrt{\sqrt{3} + 2} + 2}}{2} . And the latter one is the solution.

Hey, thanks! I finally did something right! XD Cool problem, by the way.

Drex Beckman - 5 years, 5 months ago

Did exactly same :-)

Atul Shivam - 5 years, 5 months ago

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