Given #costheta=-4/5# and #-270<theta<-180#, how do you find #sin (theta/2)#?

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Answer 1 · 2016-06-23T13:01:39

#sin(theta/2)=3/sqrt10#

Let #A=theta/2#, as #cos2A=1-2sin^2A#, we have

#costheta=1-2sin^2(theta/2)#

or #-4/5=1-2sin^2(theta/2)#

or #2sin^2(theta/2)=1+4/5=9/5#

or #sin^2(theta/2)=9/10#

As #-270^o < theta < -180^o#, #theta# is in second quadrant and is positive

Hence #sin(theta/2)=sqrt(9/10)=3/sqrt10#