Given #sintheta=-8/17# and #180<theta<270#, how do you find #cos(theta/2)#?

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Answer 1 · 2016-08-31T14:34:45

# cos (theta/2)=-1/sqrt17#.

#180<,theta<,270#.

# rArr cos theta<0, and, :., cos theta=-sqrt(1-sin^2 theta),#

where #sin theta=-8/17#

#:. cos theta=-sqrt(1-(-8/17)^2)=-15/17#.

# "Now", cos theta=2cos^2 (theta/2)-1 rArr -15/17=2cos^2 (theta/2)-1#

#rArr 2cos^2 (theta/2)=1-15/17=2/17#

# rArr cos^2 (theta/2)=1/17#

#rArr cos (theta/2)=+-1/sqrt17#

# "But", 180<,theta<,270 rArr 180/2<,theta/2<,270/2,#

i.e., #cos (theta/2)<0#.

Therefore, # cos (theta/2)=-1/sqrt17#.

Enjoy Maths.!