let's use
cos 2A = 2cos^2 A - 1
or
cos x = 2cos^2(x/2) - 1
1/4 = 2cos^2(x/2) - 1
5/8 = cos^2(x/2)
cosx = √5/√8 or √10/4
Please help me solve this problem. I have a test tomorrow and I do not understand at all.
Determine all solutions of the equations in radians.
Find cos x/2, given that cos x = 1/4 and x terminates in 0 < x < π/2.
3 answers
How did you get 5/8? Thanks for your help!
from
1/4 = 2cos^2(x/2) - 1
1/ 4 + 1 = 2cos^2(x/2)
5/4 = 2cos^2(x/2) , now divide by 2
5/8 = cos^2(x/2) , now take √
cos x/2 = √5/√8 , rationalizing
cos x/2 = (√5/√8)(√8/√8) = √40/8 = 2√10/8 = √10/4
1/4 = 2cos^2(x/2) - 1
1/ 4 + 1 = 2cos^2(x/2)
5/4 = 2cos^2(x/2) , now divide by 2
5/8 = cos^2(x/2) , now take √
cos x/2 = √5/√8 , rationalizing
cos x/2 = (√5/√8)(√8/√8) = √40/8 = 2√10/8 = √10/4
1 answer