Question

найдите: 9*√2*sin2x, если sinx=1/3; -п/2

Answer 1

sin2x=2sinxcosx
cosx=√(1-(sinx)^2)
cosx=√(1-1/9)=√(8/9)=2√2/3  тк cosx>0  при -pi/2<x<pi/2    
 9√2sin2x=9√2*21/3*2√2/3=8

Answer 2

сosx=√(1-sin²x)=√(1-1/9)=√(8/9)=2√2/3
sin2x=2sinxcosx=2*1/3*2√2/3=4√2/9
9√2sin2x=9√2*4√2/9=8