Express-2-sin-cos-6-in-the-form-sin-A-sin-B-i-using-that-result-prove-that-2sin-cos-6-cos-4-cos-2-sin-7-sin-ii-deduce-the-result-cos-12pi-7-cos-8pi-7-cos-4pi

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