Given #sin theta =-3/5# and #270<theta<360#, how do you find #sin (theta/2)#?

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Answer 1 · 2016-09-20T12:30:27

#sin(theta/2)=sqrt(1/10)#

As #sintheta=-3/5# and #270^o < theta < 360^o#

and #135^o < theta < 180^o#, and #theta/2# lies in second quadrant and #sin(theta/2)# wil be positive.

#costheta=sqrt(1-sin^2theta)=sqrt(1-(-3/5)^2)#

= #sqrt(1-9/25)=sqrt(16/25)=4/5# note As #theta# is in fourth quadrant, #costheta# is positive.

Now as #cos2A=1-2sin^2A#, #sinA=sqrt((1-cos2A)/2)#

Hence, #sin(theta/2)=sqrt((1-costheta)/2)#

= #sqrt((1-4/5)/2)=sqrt(1/10)#