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

2 Answers
Jul 27, 2018

x=1.13

Explanation:

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

Jul 27, 2018

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