# cosine of 1/6 ? r

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So I want to deduce the adjacent side length of a hypotenuse at a ?/6 angle  from the x axis. Basic stuff, right?

if

1/8 of a circumference is to its adjacent side

?/4

___

?(1/2)

as 1/12 of a circumference is to its adjacent side:

1/6?

____

x

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If the hypotenuse has length 1, the adjacent side at an ?/6 angle has length cos(?/6) = sqrt(3)/2. Is that what you wanted to know? Or do you want to know why it's sqrt(3)/2?

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If the hypotenuse has length 1, the adjacent side at an ?/6 angle has length cos(?/6) = sqrt(3)/2. Is that what you wanted to know? Or do you want to know why it's sqrt(3)/2?

Exactly, I haven't slept well so I am having trouble deriving it.

You can use pitagoras theorem for a ?/4 angle since both sides are equal at that angle, hence:

h2 = x2 + x2

x = ?(1/2)

Edited by gasto

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Draw an equilateral triangle in the usual position (one horizontal side at the bottom of the figure), then draw the vertical line that passes through the top vertex. This divides the equilateral triangle in two triangles like the one you are interested in. From this construction, it's obvious that the length of the other side is 1/2. Now use the Pythagorean theorem to compute the side you are interested in.

I get 2(?(5/4))

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I get 2(?(5/4))

You are doing it wrong. :)

Do you agree that if you split an equilateral triangle with side 1 in two by drawing a line that passes through one vertex and is perpendicular to the opposite side, the opposite side is divided in two parts of size 1/2? Now look at only one of the two triangles that we have formed. It is a right triangle with hypotenuse of length 1 and ?/6 is one of its angles. The side opposite the ?/6 angle has length 1/2. The remaining side has length x such that

x^2 + (1/2)^2 = 1^2

That's where you get x = sqrt(3)/2.

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Ah basic euclidean geometry. I must be declining in intellect with these overnight refluxes... thanks to esomeprazol and sucralfate, it is improving.

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