Given #tantheta=5/12# and #pi<theta<(3pi)/2#, how do you find #cos2theta#?

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Answer 1 · 2016-10-28T16:43:07

#cos2theta=119/169#

Parametric formula for #cosalpha#
#cosalpha=(1-t^2)/(1+t^2)# where #t=tan(alpha/2)#

#2theta=alpha, theta=alpha/2#

#cos2theta=(1-(5/12)^2)/(1+(5/12)^2)=119/169#

#5/12 < 1\ \ \ # so #\ \ \ pi < theta < 5/4pi#

#2pi < 2theta <5/2pi#

#0 < cos2theta < 1 #

#119/169# is acceptable