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I was solving a question and got struck in a sub-question which is,

If $$x=\cos1°\cos2°\cos3°\cdots\cos89°$$ Find the value of $x$.

I don't actually have any idea on how to proceed on this question. Any hints are welcome. The answer given in the book is a perfect integer $17$

Arjun
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2 Answers2

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I tried the method suggested in the comments, but it seems to bog down.

$$\begin{align} I&=\prod_{k=1}^{89}\cos{k^\circ}\\ &=\cos{45^\circ}\prod_{k=1}^{44}\cos{k^\circ}\prod_{k=46}^{89}\cos{k^\circ}\\ &=\frac1{\sqrt2}\prod_{k=1}^{44}\cos{k^\circ}\cos(90-k)^\circ\\ &=\frac1{\sqrt2}\prod_{k=1}^{44}\frac12(\cos90^\circ+\cos(90-2k)^\circ)\\ &=\frac1{2^{44}\sqrt2}\prod_{k=1}^{44}\cos(90-2k)^\circ\\ &=\frac1{2^{44}\sqrt2}\prod_{k=1}^{22}\cos(90-2k)^\circ\prod_{k=23}^{44}\cos(90-2k)^\circ\\ &=\frac1{2^{44}\sqrt2}\prod_{k=1}^{22}\frac12(\cos90^\circ+\cos(90-4k)^\circ)\\ &=\frac1{2^{66}\sqrt2}\prod_{k=1}^{22}\cos(90-4k)^\circ \end{align}$$ and now we can no longer group the arguments in pairs that add up to $90^\circ.$ At the next stage, if I follow the strategy of grouping the largest arguments with the smallest, I get $$ \frac1{2^{77}\sqrt2}\prod_{k=1}^{11}(\cos88^\circ+\cos(92-8k)^\circ),$$ which is better than what we started with, but still unpleasant.

I think it might be better to approximate the product as follows.

$$\begin{align} \log I&=\sum_{k=0}^{90}\log(\cos k^\circ)\\ &\approx \int_0^{90}\log\cos\left(\frac{\pi x}{180}\right)\mathrm{d}x\\ &=\frac{180}{\pi}\int_0^{\pi/2}\log\cos u\mathrm{d}u \end{align}$$

It is known that $$\int_0^{\pi/2}\log\cos u\mathrm{d}u=-\frac{\pi\log2}{2},$$ so we get $$\log I \approx -90\log2$$ and $$I\approx2^{-90}.$$

One can use the known error estimate in the trapezoidal rule to bound the error in this estimate, but I haven't done that.

NOTES

This may not be a very good estimate, at least in terms of relative error. A simple-minded python script gives $$I\approx 2^{-85.75}.$$ Also, if we're going to approximate it, we could just say $\cos88^\circ\approx0$ and forge ahead.

P.S.

Whoops. The derivative of $\log\cos x$ is $-\tan x$, so bounding the error with the trapezoidal rule, or the Euler-Maclaurin formula won't be easy.

saulspatz
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    Full disclosure, I didn't look any further into the suggestion I provided; I just happened to recall this being a usual "trick" for finding identities involving products of sines/cosines (case in point being the linked duplicate post). Had I done the legwork myself, I would have gotten to$$\frac1{2^{44}\sqrt2}\prod_{k=1}^{44}\cos\frac{2k\pi}{180}=\frac1{2^{44}\sqrt2}\prod_{k=1}^{44}\cos\frac{k\pi}{90}$$then consider looking for some way to condense the product of the first $44$ $90^{\rm th}$ roots of unity. – user170231 Feb 11 '20 at 19:41
  • Actually the answer is 17 – Arjun Feb 12 '20 at 04:06
  • Please vote to reopen this question, I have found a marvellous answer for it – Arjun Feb 12 '20 at 04:07
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HINT.-One has $$x=\cos1°\cos2°\cos3°\cdots\cos89°\\ \frac{2x}{\sqrt2}=(\cos1°\cos2°\cdots \cos44°)(\cos46°\cos47°\cdots\cos89^o)\space\space..... (88\text{ factors})$$ and because of $\cos(x)\cos(90-x)=\dfrac{\sin(2x)}{2}$ $$\frac{2^{45}x}{\sqrt2}=(\sin2^o\sin4^o\sin6^o\cdots\sin88^o)\space\space..... (44\text{ factors})$$ Again, because of $\sin(x)\sin(90-x)=\dfrac{\sin(2x)}{2}$ $$\frac{2^{45+22 }x}{\sqrt2}=(\sin4^o\sin8^o\sin12^o\cdots\sin84^o\sin88^o)\space\space..... (22\text{ factors})$$There must certainly be several elementary ways (without the use of Multiplier Angle Formulas) reducing the number of factors. For example we can reduce to $11$ factors to determine with calculator as follows:

$$\sin4^o\cos2^o\approx0.0697\\\sin8^o\cos6^o\approx0.1384\\ \sin12^o\cos10^o\approx0.2047\\\sin16^o\cos14^o\approx0.2674\\\sin20^o\cos18^o\approx0.3252\\\sin24^o\cos22^o\approx0.3771\\\sin28^o\cos26^o\approx0.4219\\\sin32^o\cos30^o\approx0.4589\\\sin36^o\cos34^o\approx0.4872\\\sin40^o\cos38^o\approx0.5065\\\sin44^o\cos42^o\approx0.5162$$Thus we have $$x\approx\dfrac{0.00155969\sqrt2}{2^{67}}$$

Piquito
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