Question

PLS ANYONE ANSWER THIS!!!!!!!!

Answer 1

Answer:

2

K= 1/8, solve for k using trigonometric identities

Answer 2

Answer:

K = sin(π/18). sin(5π/18). sin(7π/18)

= (1/2)[cos((5π/18)-(π/18))-cos((5π/18)+(π/18))].sin7π/18

= (1/2)[cos2π/9-cosπ/3].sin7π/18

= (1/2)[cos2π/9.sin7π/18]-(1/4)sin7π/18

= (1/4)[sin((2π/9)+(7π/18))+sin((7π/18)-(2π/9)]-(1/4)sin7π/18

= (1/4)[sin11π/18+sinπ/6]-(1/4)sin7π/18

= (1/4)sin11π/18+(1/8)-(1/4)sin7π/18

= (1/4)sin(π-(11π/18))+(1/8)-(1/4)sin7π/18

= (1/4)sin7π/18+(1/8)-(1/4)sin7π/18

= 1/8 i.e (1/2√2)^2

therefore, log1/2√2 [(1/2√2)^2] = 2

Hence, answer is 2.