If cos(α)=4/5 and we know that α∈(270˚; 360˚), how much is cos(α/2)?

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Answer 1 · 2017-11-03T22:09:19

#cos(alpha/2)=-(3sqrt(10))/10#

Using the identity:

#cos^2(x/2)= 1/2(1+cosx)#

#cos(alpha/2)=sqrt(1+cosalpha)/sqrt(2)#

If #cosalpha=4/5#

#:.#

#cos(alpha/2)=(sqrt(1+4/5))/sqrt(2)=(sqrt(9/5))/sqrt(2)=(3/sqrt(5))/sqrt(2)=-(3sqrt(10))/10#

Note:

Cosine is negative because #alpha/2# is in the III quadrant and #alpha# is in the IV quadrant.