Given cos(t)=-7/9 and sin(t)= -sqrt32/9, both in the 3rd Quadrant how do you find cos(t/2) and sin(t/2)?

1 Answer
Jan 12, 2017

#cos(t/2)=-1/3#
#sin(t/2)=(2sqrt2)/3#

Explanation:

A:
#cos(t)=cos(t/2+t/2)#
#-7//9=cos^2(t/2)-sin^2(t/2)#
#-7/9=cos^2(t/2)-(1-cos^2(t/2))#
#-7/9=2cos^2(t/2)-1#
#-7/9+1=2cos^2(t/2)#
#2/9=2cos^2(t/2)#
#1/9=cos^2(t/2)#
#+-sqrt(1/9)=+-1/3=cos(t/2)#
#cos(t/2)=-1/3# , it has a -ve value because t/2 lie at 2nd quadrant.

B:
#sin(t)=sin(t/2+t/2)#
#-sqrt32/9=2sin(t/2)cos(t/2)#
#-sqrt32/9=2sin(t/2)(-1/3)#, from answer above #cos(t/2)=-1/3#
#-sqrt32/9=-2/3sin(t/2)#
#-sqrt32/9*-3/2=sin(t/2)#
#-sqrt(16*2)/9*-3/2=sin(t/2)#
#-4sqrt(2)/9*-3/2=sin(t/2)#
#(2sqrt2)/3=sin(t/2)#