How do you find csc ø/2 if sin ø = 8/17 and π/2 < ø < π ??

1 Answer
Mar 28, 2018

# csc (theta/2)=sqrt17/4#

Explanation:

#sin theta = 8/17 ; pi/2< theta< pi :. theta # is in 2nd quadrant

#Sin theta = P/H ; P=8 , H=17 :. B^2= H^2-P^2= 17^2-8^2#

#:. B^2=225 :. B=15 , Cos theta = -B/H= -15/17 ; cos theta#

is negative in 2nd quadrant.

# sin theta=2sin (theta/2) cos (theta/2)=8/17#

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

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

#cos^2(theta/ 2)=1/17:. cos(theta/ 2)=1/sqrt17 :.#

# 2sin (theta/2) cos (theta/2)=8/17# or

# 2sin (theta/2)*1/sqrt17=8/17# or

# 2sin (theta/2)=8/17*sqrt17# or

# sin (theta/2)=4/17*sqrt17= 4/sqrt17#

#:. csc (theta/2)=1/sin (theta/2)=sqrt17/4# [Ans]