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Logarithmic Differentiation

2 cos θ - 1...

Question

$(2 c o s θ - 1) (2 c o s 2 θ - 1) (2 c o s 2^{2} θ - 1) . . . . . . (2 c o s 2^{n - 1} θ - 1) = \frac{2 c o s 2^{n} θ + 1}{2 c o s θ + 1}$

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\frac{2 cos 2^{n} θ + 1}{2 cos θ + 1} = (2 cos θ - 1) (2 cos 2 θ - 1) (2 cos 2^{2} θ - 1) \dots (2 cos 2^{n - 1} θ - 1)

f_{n} (θ) = (2 cos θ + 1) (2 cos θ - 1) (2 cos 2 θ - 1) (2 cos (2^{2}) θ - 1) . . . . . . (2 cos (2^{n - 1}) θ - 1)

1 +^{n} C_{1} cos θ +^{n} C_{2} cos 2 θ + . . . . . . . . . +^{n} C_{n} cos n θ

2 c o s θ = x = \frac{1}{x} a n d 2 c o s ϕ = y + \frac{1}{y}

x y z . . . . + \frac{1}{x y z . . . .} = 2 c o s (θ + ϕ + . . . .)

2 c o s θ = x = \frac{1}{x} a n d 2 c o s ϕ = y + \frac{1}{y}

\frac{x}{y} + \frac{y}{x} = 2 c o s (θ - ϕ)

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