Question

show that cos(2tan inverse 1/7) = sin(4tan inverse1/3)

Answer 1

tan α = 1/7 
sin α = 1/√(1²+7²) = 1/√50 
cos α = 7/√(1²+7²) = 7/√50 

β = tan⁻¹(1/3) 
tan β = 1/3 
sin β = 1/√(1²+3²) = 1/√10 
cos β = 3/√(1²+3²) = 3/√10 

cos(2 tan⁻¹(1/7)) 
= cos(2α) 
= cos²α − sin²α 
= (7/√50)² − (1/√50)² 
= 49/50 − 1/50 
= 24/25 

sin(4 tan⁻¹(1/3)) 
= cos(4β) 
= 2 sin(2β) cos(2β) 
= 2 (2 sinβ cosβ) (cos²β − sin²β) 
= 4 (1/√10) (3/√10) (9/10 − 1/10) 
= 12/10 * 8/10 
= 24/25