If - - -2n - 1- prove that 2n cos- cos 2- cos 22 -cos 2n - 1 - - 1

L.H.S = 2^n · sin 2^n θ2^n sin θ where θ = π2^n + 1 = [ sin2^n π2^n + 1 ]÷ ( sinπ2^n + 1 ) = sin(π α) + si ...

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Domain and Range of Trigonometric Ratios

If θ = π2n ...

Question

If $θ = \frac{π}{2^{n} + 1}$ , prove that
$2^{n} cos θ cos 2 θ cos 2^{2} θ . . . . cos 2^{n - 1} θ = 1$

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Solution

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(2^{n} + 1) θ = π,

2^{n} \cos θ \cos 2 θ \cos 2^{2} θ . . . \cos 2^{n - 1} θ = 1

- 1

θ = \frac{π}{2^{n} + 1}

cos θ cos 2 θ 2 cos 2^{2} θ . . . cos 2^{n - 1} θ = \frac{1}{a^{n}}

a

cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{4 π}{7} = - \frac{1}{8}

cos θ cos 2 θ cos 4 θ . . . cos 2^{n - 1} θ = - \frac{1}{2^{n}}

θ = \frac{π}{2^{n} - 1}

cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{4 π}{7} = \frac{1}{8}

cos θ cos 2 θ cos 4 θ . . . cos 2^{n - 1} θ = \frac{1}{2^{n}}

θ = \frac{π}{2^{n} - 1}

c o s θ . c o s 2 θ . c o s 2^{2} θ . c o s 2^{3} θ \dots c o s 2^{n - 1} θ = \frac{s i n 2^{n} θ}{2^{n} s i n θ}

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