Question #ec3cf

1 Answer
Jan 15, 2018

Given

#180 < x < 270#

#=>180/2 < x/2 < 270/2#

#=>90 < x/2 < 135#

So #sin(x/2)->"positive"#

And #cos(x/2)->"negative"#

Now #cos(x)=-4/5 #

#=>2cos^2(x/2)-1=-4/5 #

#=>2cos^2(x/2)=1-4/5 #

#=>2cos^2(x/2)=1/5 #

#=>cos^2(x/2)=1/10 #

#=>cos(x/2)=-1/sqrt10 #

Again
#cos(x)=-4/5 #

#=>1-2sin^2(x/2)=-4/5 #

#=>2sin^2(x/2)=1+4/5 #

#=>2sin^2(x/2)=9/5 #

#=>sin^2(x/2)=9/10 #

#=>sin(x/2)=3/sqrt10 #