How do you find the exact values of #sin2u, cos2u, tan2u# using the double angle values given #secu=-5/2, pi/2<u<pi#?

1 Answer

#sin 2u = - (4sqrt21)/25#
#cos 2u = - 17/25#
#tan 2u = (4sqrt21)/17#

Explanation:

#sec u = 1/(cos u) = - 5/2# --> #cos u = - 2/5#.
Find #sin u#
#sin ^2 u = 1 - cos^2 u = 1 - 4/25 = 21/25#
#sin u = +- sqrt21/5#.
Since #u# is in Quadrant II, then #sinu# is positive.

#sin2u=2sinucosu=2(sqrt21/5)(-2/5) =- (4sqrt21)/25#

#cos 2u = 2cos^2 u - 1 = 2(4/25) - 1 = - 17/25#
#tan 2u = sinu/cosu = (- (4sqrt21)/25)(- 25/17) = (4sqrt21)/17#