Solution: Since sina=4/5 and A belongs to (0, π/2) = >; cosa & gt0
Sin 2a+cos 2a = 1 = > Cosa= radical sign [1-(4/5) 2] = 3/5 (positive values only).
sin2a = 2 * Sina * cosa =(3/5)*(4/5)= 12/25
cos2a = 1-2sin^2a=2cos^2a=9/25
tan2a =(sin2a)/(cos2a)=( 12/25)/(9/25)= 4/3
Summary: this question belongs to the basic concept, the memory of special formulas of trigonometric functions:
Sin 2A+COS 2A =1sia2a = 2 Sinacosa COS2A = COS 2A-SIN 2A = 2 COS 2A-1=1-2 SIN 2A
tan a = Sina/cosa sin(a+b)= Sina * cosb+cosa * sinb cos(a+b)= cosa cosb-Sina * sinb
sin(a-b)= Sina * cosb-cosa * sinb cos(a-b)= cosa * cosb+Sina sinb
These are commonly used, and a few are not written here. I hope you can master these formulas! ! ! ! !