Given #cos(x) = -3/-7# and #pi/2 < x < pi#, how do you find #cos(x/2)#?

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Answer 1 · 2018-07-27T10:52:04

#x=1.13#

#3/7=cosx#

#=cos(2timesx/2)#

#=cos^2(x/2)-sin^2(x/2)#

#=cos^2(x/2)-(1-cos^2(x/2))#

#=2cos^2(x/2)-1#

#3/7=2cos^2(x/2)-1#

#2cos^2(x/2)=10/7#

#cos^2(x/2)=5/7#

#cos(x/2)=+-sqrt(5/7)#

The domain was originally #pi/2 < x < pi# but since it is now #x/2#, the domain has changed to #pi/4 < x/2 < pi/2# ie first quadrant only

Therefore, #cos(x/2)=sqrt(5/7)# only as cosine is positive in the first and fourth quadrant and negative in the second and third quadrant

#x/2=0.56#

#x=1.13#

Answer 2 · 2018-07-27T12:57:04

#cos x = -3 / ( - 7 ) = 3/7 rArr x in Q_1# or #Q_4#.

So, #pi/2 < x < pi # is incompatible.

If #cos x = - 3/7, x in Q_2# or #Q_3 and x/2 in Q_1 #or Q_2#.

#cos ( x/2 ) = sqrt (1/2(1+ cos x )) = sqrt (2/7), if x in Q_2#, and
#= - sqrt (2/7), x in Q_3#.

Additional examples:

If #x = 2/3pi, cos x = -1/2# and # cos (x/2) = cos (pi/3) = 1/2#

If #x = 4/3pi, cos x = -1/2# and

# cos (x/2 ) = cos (2/3pi) = - 1/2#