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]