Question Number 101040 by Rio Michael last updated on 30/Jun/20

 Express 2 sin θ cos 6θ in the form  sin A − sin B  (i) using that result prove that 2sin θ( cos 6θ + cos 4θ + cos 2θ) = sin 7θ−sin θ  (ii) deduce the result cos (((12π)/7)) + cos (((8π)/7)) + cos (((4π)/7)) = −(1/2)  (iii) hence find a general solution to ((sin7θ − sin θ)/(cos 6θ + cos 4θ + cos 2θ)) = 1

Answered by maths mind last updated on 30/Jun/20

2sin(θ)cos(6θ)=sin(θ+6θ)+sin(θ−6θ)=sin(7θ)+sin(−5θ)  2sin(θ)cos(4θ)=sin(5θ)+sin(−3θ)  2sin(θ)cos(2θ)=sin(3θ)+sin(−θ)  2sin(θ)(cos(6θ)+cos(4θ)+cos(2θ))=sin(7θ)−sin(5θ)+sin(5θ)−sin(3θ)  +sin(3θ)−sin(θ)=sin(7θ)−sin(θ)  θ=((2π)/7)  ⇒2sin(((2π)/7))(cos(((4π)/7))+cos(((8π)/7))+cos(((4π)/7)))=sin(2π)−sin(((2π)/7)))  ⇔cos(((4π)/7))+cos(((8π)/7))+cos(((4π)/7))=−(1/2)  (iii)⇔sin(7θ)−sin(θ)=cos(2θ)+cos(4θ)+vos(2θ)  ⇔2sin(θ)=1⇒sin(θ)=(1/2)⇒2kπ+(π/6),2kπ+((5π)/6),k∈Z

Commented byRio Michael last updated on 30/Jun/20

explicit sir thanks