precalculus
posted by lina
use a sum or difference formula to find the exact of the trigonometric function
cos(13pi/12)

Steve
cos(13pi/12)
= cos(13pi/12 + 24pi/12)
= cos(11pi/12)
= cos(pi  11pi/12)
= cos(pi/12)
now, cos(x/2) = √((1+cos(x))/2)
cos(pi/12) = √((1+cos(pi/6))/2)
= √((1+√3/2)/2)
= √(2 + √3/4)
= 1/2 √(2+√3)
or
1/2√2 (1+√3)
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