Asked by lina
use a sum or difference formula to find the exact of the trigonometric function
cos(-13pi/12)
cos(-13pi/12)
Answers
Answered by
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)
= 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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