If #costheta=4/5# and #(3pi)/2<theta<2pi#, how do you find #cos (theta/2)#?

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Answer 1 · 2016-09-20T11:32:18

#cos(theta/2)=-sqrt(9/10)#

Using the identity #cos2A=2cos^2A-1#, we have

#costheta=2cos^2(theta/2)-1# and hence #cos(theta/2)=sqrt((1+costheta)/2)#

As #costheta=4/5#

#cos(theta/2)=+-sqrt((1+4/5)/2)#

= #+-sqrt((9/5)/2)=+-sqrt(9/10)#

Now as #(3pi)/2 < theta<2pi#,

#(3pi)/4 < theta/2 < pi# i.e. #theta/2# lies in second quadrant and

#cos(theta/2)=-sqrt(9/10)#