If #sintheta=sqrt403/22# and #pi/2<theta<pi#, how do you find #tan2theta#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Hubert Nov 20, 2016 #tan2theta=(9sqrt403)/161# Explanation: If #theta in [pi/2,pi]#, #costheta<0#, so: #costheta=-sqrt{1-sin^2theta}# #costheta=-sqrt{1-403/484}=-sqrt{81/484}=-9/22# #tantheta=sintheta/costheta=(sqrt{403}/22)/(-9/22)=-sqrt403/9# #tan2theta=(2tantheta)/(1-tan^2theta)=(-(2sqrt403)/9)/(1-403/81)=(-(2sqrt403)/9)/(-322/81)=(9sqrt403)/161# Answer link Related questions What are Double Angle Identities? How do you use a double angle identity to find the exact value of each expression? How do you use a double-angle identity to find the exact value of sin 120°? How do you use double angle identities to solve equations? How do you find all solutions for #sin 2x = cos x# for the interval #[0,2pi]#? How do you find all solutions for #4sinthetacostheta=sqrt(3)# for the interval #[0,2pi]#? How do you simplify #cosx(2sinx + cosx)-sin^2x#? If #tan x = 0.3#, then how do you find tan 2x? If #sin x= 5/3#, what is the sin 2x equal to? How do you prove #cos2A = 2cos^2 A - 1#? See all questions in Double Angle Identities Impact of this question 2023 views around the world You can reuse this answer Creative Commons License