# Re: pre-calc-is this the right way????

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Posted by Jonathan Burros on October 23, 2002 at 16:31:52:

In Reply to: pre-calc-is this the right way???? posted by Jacquie on October 21, 2002 at 18:17:29:

: find exact value of cos 11pi/12
: i used the double-angle formula for cos
: sq. root of (1 + cos 11pi/12)/2
: sq. root of(1+cos(8pi/12 + 3pi/12))/2
: sq. root of(1+cos(2pi/3 + pi/4))/2
: sq. root of(1+(-1/2)+ (sq.root of2/2))/2
: sq. root of(2/2+sq. root of2-1/2/2
: sq.root of(sq. root of 2 +1/2)/2
: but i dont know how to further reduce this???

cos 11pi/12 = cos[(12pi - pi)/12]

= cos(pi - pi/12)

= cos pi cos(pi/12) + sin pi sin(pi/12)

= - cos(pi/12)

since the angle pi/12 is in the first quadrant,
its cosine is positive.

So cos 11 pi/12 is negative

Let x = cos 11pi/12

x^2 = cos^2 11 pi/12

= (1 + cos 11 pi/6)/2

2x^2 - 1 = cos 11 pi/6

= cos[(12 pi - pi)/6]

= cos(2 pi - pi/6)

= cos pi/6

= sqrt 3/2

Solve for x and use the negative root

2x^2 = (2 + sqrt 3)/2

x^2 = (2 + sqrt 3)/4

x = - (1/2) sqrt[2 + sqrt 3]

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