trig product (1 Viewer)

juantheron · May 19, 2012

$\hspace{-16}$Can we calculate the product of\\\\\\ $\bf{\cot(1^0).\cot(2^0).\cot(3^0)..........\cot(44^0)=}$\\\\\\ Where all angle are in Degree.$

like by using Trigonometry OR Using Complex no.

can anyone help me to solve the Given Question

Thanks

fortune_cookie · May 19, 2012

http://en.wikipedia.org/wiki/Morrie's_law

Carrotsticks · May 19, 2012

juantheron said:
$\hspace{-16}$Can we calculate the product of\\\\\\ $\bf{\cot(1^0).\cot(2^0).\cot(3^0)..........\cot(44^0)=}$\\\\\\ Where all angle are in Degree.$

like by using Trigonometry OR Using Complex no.

can anyone help me to solve the Given Question

Thanks

A couple things to notice:

1. We can pair them up so the it becomes:

$\cot(1) \cot(44) \times \cot(2) \cot(43) \times \cot(3) \cot(42) \times ... \cot(k) \cot (45 - k)$

2. If we convert to Radians, we have the following expression:

$\prod_{k} \cot (\frac{k \pi}{180}) \cot(\frac{\pi}{4} - \frac{k \pi}{180})$

3. Consider the identity:

$\prod_{k=1}^{n-1} \cot \frac{k \pi}{n} = \frac{sin \frac{ n \pi }{2}}{n}$

This should be enough to get you started...

fortune_cookie said:
http://en.wikipedia.org/wiki/Morrie's_law

Good thinking, but Morrie's Law only works if the angle (inside of bracket) are multiples each other by a scale of 2.

RealiseNothing · May 19, 2012

I'm trying something, it may or may not work out.

Using what Carrot posted: $\prod_{k=1}^{n-1} \cot \frac{k \pi}{n} = \frac{2^{n-1}}{sin \frac{ n \pi }{2}}$

$\prod_{k=1}^{44} \cot \frac{\frac{k}{4} \pi}{45} = \frac{2^{44}}{sin \frac{ 45 \pi }{2}}$

$\prod_{k=1}^{44} \cot \frac{\frac{\pi}{4}(45-k)}{45} = \frac{2^{44}}{sin \frac{ 45 \pi }{2}}$

I just did some re-arranging to get it into the form we wanted. Although I'm not entirely sure if I can do what I did..... Anyway:

So we now have:

$\frac{2^{44}}{sin \frac{ 45 \pi }{2}} \times \frac{2^{44}}{sin \frac{ 45 \pi }{2}}$

$=\frac{2^{88}}{sin^2\frac{45\pi}{2}}$

This is probably wrong but whatever, some one else might get an idea from looking at this.

juantheron · May 19, 2012

Thanks Moderator But when i have searched Then I have found following values of product of sin and cos , which is Given below

$\hspace{-16}\bf{\prod_{k=1}^{n-1}\sin \left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}}$\\\\\\ \underline{\underline{Proof}}\;\; ::\;\; Using Complex no.\\\\\\ $\bf{z^n-1=(z-1).(z-z_{1}).(z-z_{2})........(z-z_{n-1})}$\\\\\\ Where $\bf{z}$ is a Complex no. and $\bf{z_{k}=\left(\cos (k.\alpha)+i.\sin(k.\alpha)\right)}$ and $\bf{\alpha=\frac{2\pi}{n}}$\\\\\\ Now $\bf{\frac{z^n-1}{z-1}=(z-z_{1}).(z-z_{2})........(z-z_{n-1})}$\\\\\\ $\bf{\lim_{z\rightarrow 1}\frac{z^n-1}{z-1}=\lim_{z\rightarrow 1}(z-z_{1}).(z-z_{2})........(z-z_{n-1})}$\\\\\\ So $\bf{n=(1-\cos(\alpha)-i.\sin(\alpha)).(1-\cos(2\alpha)-i.\sin(2\alpha))......(1-\cos((n-1)\alpha)-i.\sin((n-1)\alpha))}$\\\\\\ Taking Modulus on both side, We Get\\\\\\ $\bf{n=2^{n-1}.\sin \left(\frac{\alpha}{2}\right).\sin \left(\frac{2\alpha}{2}\right).....\sin \left(\frac{(n-1)\alpha}{2}\right)}$\\\\\\ Let $\bf{\frac{\alpha}{2}=\frac{\pi}{n}}$\\\\\\$

$\hspace{-16}$We Get $\bf{n=2^{n-1}.\prod_{k=1}^{n}\sin \left(\frac{k\pi}{n}\right)}$\\\\\\ $\bf{\prod_{k=1}^{n-1}\sin \left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}}$\\\\\\ Similarly $\bf{\prod_{k=1}^{n-1}\cos \left(\frac{k\pi}{n}\right)=\frac{\sin\left(\frac{n\pi}{2}\right)}{2^{n-1}}}$\\\\\\ \underline{\underline{proof}}:: Using Complex No.\\\\ $\bf{z^n+1=(z+1).(z+z_{1}).(z+z_{2})............(z+z_{n-1})}$\\\\\\ Where $\bf{z}$ is a Complex no. and $\bf{z=\cos(k\alpha)+i.\sin(k\alpha)}$ and $\bf{\alpha=\frac{2\pi}{n}}$\\\\\\ Now $\bf{\frac{z^n+1}{z+1}=(z+z_{1}).(z+z_{2})........(z+z_{n-1})}$\\\\\\ $\bf{\lim_{z\rightarrow 1}\frac{z^n+1}{z+1}=\lim_{z\rightarrow 1}(z+z_{1}).(z+z_{2})........(z+z_{n-1})}$\\\\\\ So $\bf{1=(1+\cos(\alpha)+i.\sin(\alpha)).(1+\cos(2\alpha)+i.\sin(2\alpha))......(1+\cos((n-1)\alpha)+i.\sin((n-1)\alpha))}$\\\\\\ taking Modulus on both side, We Get\\\\\\$

$\hspace{-16}\bf{1=2^{n-1}.\cos \left(\frac{\alpha}{2}\right).\cos \left(\frac{2\alpha}{2}\right).....\cos \left(\frac{(n-1)\alpha}{2}\right)}$\\\\\\ So $\bf{\prod_{k=1}^{n-1}\cos \left(\frac{k\pi}{n}\right)=\frac{1}{2^{n-1}}}$\\\\\\ So $\bf{\prod_{k=1}^{n-1}\cot \left(\frac{k\pi}{n}\right)=\frac{1}{n}}$\\\\\\$

which is True for all odd positive Integers, and for even it is = 0

but I did not understand how can i get $\sin \left(\frac{\pi.n}{2}\right)$

So Where I have make mistake plz verify it

Thanks

Carrotsticks · May 19, 2012

Oh yes, juantheron I made a typo. I mistyped the expression (got cos and cot mixed up).

$\prod_{k=1}^{n-1} \cot \frac{k \pi }{n} = \frac{\sin \frac{\pi n}{2}}{n}$

You acquire this by getting the product with cos, then putting it over the product with sine.

juantheron · May 20, 2012

Thanks Moderator I did not Understand How can I get value of $\cot(1^0).\cot(2^0)......\cot(44^0)$ Using above formula.

Thanks

zeebobDD · May 20, 2012

lol your questions seem very out of "HSC" range where are you getting these?

Mr Slick · May 20, 2012

is it Maths Ext 2? :S

iSplicer · May 20, 2012

Mr Slick said:
is it Maths Ext 2? :S

It could technically pass as one of the upper end questions from 4u.

Excellent contributions carrotsticks, RealiseNothing and juantheron!

Mr Slick · May 20, 2012

highly doubt it

iSplicer said:
It could technically pass as one of the upper end questions from 4u.

Excellent contributions carrotsticks, RealiseNothing and juantheron!

math man · May 20, 2012

this would be the opener of my paper

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