PLS ANYONE ANSWER THIS!!!!!!!!
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Answer:
2
K= 1/8, solve for k using trigonometric identities
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1
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.
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