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


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

$Express 2 \sin θ \cos 6 θ in the form \sin A - \sin B$ $(i) using that result prove that 2 \sin θ (\cos 6 θ + \cos 4 θ + \cos 2 θ) = \sin 7 θ - \sin θ$ $(i i) deduce the result \cos (\frac{12 π}{7}) + \cos (\frac{8 π}{7}) + \cos (\frac{4 π}{7}) = - \frac{1}{2}$ $(i i i) hence find a general solution to \frac{\sin 7 θ - \sin θ}{\cos 6 θ + \cos 4 θ + \cos 2 θ} = 1$
	Answered by maths mind last updated on 30/Jun/20

	$2 s i n (θ) c o s (6 θ) = s i n (θ + 6 θ) + s i n (θ - 6 θ) = s i n (7 θ) + s i n (- 5 θ)$ $2 s i n (θ) c o s (4 θ) = s i n (5 θ) + s i n (- 3 θ)$ $2 s i n (θ) c o s (2 θ) = s i n (3 θ) + s i n (- θ)$ $2 s i n (θ) (c o s (6 θ) + c o s (4 θ) + c o s (2 θ)) = s i n (7 θ) - s i n (5 θ) + s i n (5 θ) - s i n (3 θ)$ $+ s i n (3 θ) - s i n (θ) = s i n (7 θ) - s i n (θ)$ $θ = \frac{2 π}{7}$ $\Rightarrow 2 s i n (\frac{2 π}{7}) (c o s (\frac{4 π}{7}) + c o s (\frac{8 π}{7}) + c o s (\frac{4 π}{7})) = s i n (2 π) - s i n (\frac{2 π}{7}))$ $\Leftrightarrow c o s (\frac{4 π}{7}) + c o s (\frac{8 π}{7}) + c o s (\frac{4 π}{7}) = - \frac{1}{2}$ $(i i i) \Leftrightarrow s i n (7 θ) - s i n (θ) = c o s (2 θ) + c o s (4 θ) + v o s (2 θ)$ $\Leftrightarrow 2 s i n (θ) = 1 \Rightarrow s i n (θ) = \frac{1}{2} \Rightarrow 2 k π + \frac{π}{6}, 2 k π + \frac{5 π}{6}, k \in Z$
	Commented by Rio Michael last updated on 30/Jun/20

	$explicit sir thanks$
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