Find out the answer to "Proving:cos⁡A⋅cos⁡2A⋅cos⁡22A⋅cos⁡23A…cos⁡2n−1A=sin⁡2nA2nsin⁡A"

iristh3virusoo2

Answered question

2022-02-26

Proving:

\cos A \cdot \cos 2 A \cdot {\cos 2}^{2} A \cdot {\cos 2}^{3} A \dots {\cos 2}^{n - 1} A = \frac{{\sin 2}^{n} A}{2^{n} \sin A}

vzletanje8z0

Beginner2022-02-27Added 6 answers

You can prove it by induction since

\sin t = 2 \cos \frac{t}{2} \sin \frac{t}{2}

\sin t = 2^{2} \cos \frac{t}{2} \cos \frac{t}{4} \sin \frac{t}{4}

\sin t = 2^{3} \cos \frac{t}{2} \cos \frac{t}{4} \cos \frac{t}{8} \sin \frac{t}{8}

So we conjecture:

\sin t = 2^{n} \sin \frac{t}{2^{n}} \prod_{k = 1}^{n} \cos \frac{t}{2^{k}}

It is true for

n = 1

\sin t = 2^{1} \sin \frac{t}{2^{1}} \prod_{k = 1}^{1} \cos \frac{t}{2^{1}} = 2 \cos \frac{t}{2} \sin \frac{t}{2}

But then for

n + 1

we get

\sin t = 2^{n + 1} \sin \frac{t}{2^{n + 1}} \prod_{k = 1}^{n + 1} \cos \frac{t}{2^{k}}

\sin t = 2^{n} 2 \sin \frac{t}{2^{n + 1}} \cos \frac{t}{2^{n + 1}} \cos \frac{t}{2^{n + 1}} \prod_{k = 1}^{n} \cos \frac{t}{2^{k}}

\sin t = 2^{n} \sin \frac{t}{2^{n}} \prod_{k = 1}^{n} \cos \frac{t}{2^{k}}

vazen2bl

Beginner2022-02-28Added 9 answers

I tried to prove it formally, without direct mathematical induction.

a_{n} = {\cos 2}^{n - 1} A

\Rightarrow a_{n} = \frac{2 {\sin 2}^{n - 1} A \cdot {\cos 2}^{n - 1} A}{2 {\sin 2}^{n - 1} A}

\Rightarrow a_{n} = \frac{{\sin 2}^{n} A}{2 {\sin 2}^{n - 1} A}

\Rightarrow a_{n} = \frac{V_{n + 1}}{2 V_{n}}

Where,

V_{n} = {\sin 2}^{n - 1} A

So,

a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} \dots a_{n}

= \frac{V_{2}}{2 V_{1}} \cdot \frac{V_{3}}{2 V_{2}} \cdot \frac{V_{4}}{2 V_{3}} \cdot \frac{V_{5}}{2 V_{4}} \dots \frac{V_{n + 1}}{2 V_{n}}

= \frac{V_{n + 1}}{2^{n} V_{1}}

= \frac{{\sin 2}^{n} A}{2^{n} \sin A}

Note, here

2^{m - 1} A \neq z π

, where

m \in N, n \in N, z \in Z

and

m \geq n

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