Question

The value of
product of Re[Cis(2n-1)°] from n=1 to 45
Whole upon the
product of Im[cis2(2n-1)°] from n=1 to n=45

Answer 1

@Michele_laino

Answer 2

is \[\huge cis\theta = \cos \theta + i\sin \theta \]

Answer 3

Yup

Answer 4

The question is like cos1°cos3°cos5°....cos89°/sin2(1)°sin2(3°)....sin2(89)°

Answer 5

then we have to evaluate this quantity:
\[\huge \prod\limits_{n = 1}^{45} {\frac{{\cos \left( {2n - 1} \right)}}{{\sin \left\{ {2\left( {2n - 1} \right)} \right\}}}} \]

Answer 6

What I did was first solved the denominator term by applying sin2A =2sinAcosA
And cancelled all the cos term of the denominator

Answer 7

With a degree symbol also :)

Answer 8

ok!
so we have:
\[\Large \prod\limits_{n = 1}^{45} {\frac{{\cos \left( {2n - 1} \right)^\circ }}{{\sin \left\{ {2\left( {2n - 1} \right)} \right\}^\circ }}} = \prod\limits_{n = 1}^{45} {\frac{1}{{2\sin \left( {2n - 1} \right)^\circ }}} \]

Answer 9

Correct!

Answer 10

I'm thinking...

Answer 11

N I m waiting..... lolxccc

Answer 12

Sir one more thing

Answer 13

ok!

Answer 14

If u luk at the terms carefully u will see that they are like 2sin1°2sin3°2sin5°...2sin85°2sin87°
2sin89°
That means the last term is actually 2cos1° in disguise..
And so again if we combine the two terms first and last we get 2sin2°2sin6°....

Answer 15

I view such product in a different way, here it is:

\[\begin{gathered}
\prod\limits_{n = 1}^{45} {\frac{{\cos \left( {2n - 1} \right)^\circ }}{{\sin \left\{ {2\left( {2n - 1} \right)} \right\}^\circ }}} = \prod\limits_{n = 1}^{45} {\frac{1}{{2\sin \left( {2n - 1} \right)^\circ }}} = \hfill \\
\hfill \\
= \frac{1}{{{2^2}\left( {\sin 1 \cdot \sin 89} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 3 \cdot \sin 87} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 5 \cdot \sin 84} \right)}} \cdot ... \hfill \\
\hfill \\
... \cdot \frac{1}{{2\sin 45}} \hfill \\
\end{gathered} \]

Answer 16

oops.. I have made a typo:
\[\begin{gathered}
\prod\limits_{n = 1}^{45} {\frac{{\cos \left( {2n - 1} \right)^\circ }}{{\sin \left\{ {2\left( {2n - 1} \right)} \right\}^\circ }}} = \prod\limits_{n = 1}^{45} {\frac{1}{{2\sin \left( {2n - 1} \right)^\circ }}} = \hfill \\
\hfill \\
= \frac{1}{{{2^2}\left( {\sin 1 \cdot \sin 89} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 3 \cdot \sin 87} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 5 \cdot \sin 85} \right)}} \cdot ... \hfill \\
\hfill \\
... \cdot \frac{1}{{2\sin 45}} \hfill \\
\end{gathered} \]

Answer 17

Sir, I meant the same BT forgot to write that in denominator...
Now will u also do it like that only sin89°=cos1°

Answer 18

ok!
I think to use this identity, for example:
\[\sin 1 \cdot \sin 89 = \frac{{ - 1}}{2}\left( {\cos 90 - \cos 88} \right) = \frac{{\cos 88}}{2}\]

Answer 19

Okay!

Answer 20

What next?

Answer 21

so, I can write this:
\[\begin{gathered}
\prod\limits_{n = 1}^{45} {\frac{{\cos \left( {2n - 1} \right)^\circ }}{{\sin \left\{ {2\left( {2n - 1} \right)} \right\}^\circ }}} = \prod\limits_{n = 1}^{45} {\frac{1}{{2\sin \left( {2n - 1} \right)^\circ }}} = \hfill \\
\hfill \\
= \frac{1}{{{2^2}\left( {\sin 1 \cdot \sin 89} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 3 \cdot \sin 87} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 5 \cdot \sin 85} \right)}} \cdot ... \hfill \\
\hfill \\
... \cdot \frac{1}{{2\sin 45}} = \hfill \\
\hfill \\
= \frac{1}{{2\cos 88}} \cdot \frac{1}{{2\cos 84}} \cdot \frac{1}{{2\cos 80}} \cdot ... \cdot \cdot \frac{1}{{2\cos 4}} \cdot \frac{1}{{2\sin 45}} \hfill \\
\hfill \\
\end{gathered} \]

Answer 22

Yep

Answer 23

Then!

Answer 24

I'm thinking...

Answer 25

Ok

Answer 26

next step:
\[\begin{gathered}
\prod\limits_{n = 1}^{45} {\frac{{\cos \left( {2n - 1} \right)^\circ }}{{\sin \left\{ {2\left( {2n - 1} \right)} \right\}^\circ }}} = \prod\limits_{n = 1}^{45} {\frac{1}{{2\sin \left( {2n - 1} \right)^\circ }}} = \hfill \\
\hfill \\
= \frac{1}{{{2^2}\left( {\sin 1 \cdot \sin 89} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 3 \cdot \sin 87} \right)}} \cdot \frac{1}{{{2^2}\left( {\sin 5 \cdot \sin 85} \right)}} \cdot ... \hfill \\
\hfill \\
... \cdot \frac{1}{{2\sin 45}} = \hfill \\
\hfill \\
= \frac{1}{{2\cos 88}} \cdot \frac{1}{{2\cos 84}} \cdot \frac{1}{{2\cos 80}} \cdot ... \cdot \cdot \frac{1}{{2\cos 4}} \cdot \frac{1}{{2\sin 45}} = \hfill \\
\hfill \\
= \frac{1}{{{2^2}\left( {\cos 4 \cdot \cos 64} \right)}} \cdot \frac{1}{{{2^2}\left( {\cos 8 \cdot \cos 68} \right)}} \cdot \frac{1}{{{2^2}\left( {\cos 12 \cdot \cos 72} \right)}} \cdot \hfill \\
\hfill \\
\cdot \frac{1}{{{2^2}\left( {\cos 16 \cdot \cos 76} \right)}} \cdot \frac{1}{{{2^2}\left( {\cos 20 \cdot \cos 80} \right)}} \cdot \frac{1}{{{2^2}\left( {\cos 24 \cdot \cos 84} \right)}} \cdot \hfill \\
\hfill \\
\cdot \frac{1}{{{2^2}\left( {\cos 28 \cdot \cos 88} \right)}} \cdot \hfill \\
\hfill \\
\cdot \frac{1}{{{2^2}\left( {\cos 32 \cdot \cos 60} \right)}} \cdot \frac{1}{{{2^2}\left( {\cos 36 \cdot \cos 56} \right)}} \cdot \frac{1}{{{2^2}\left( {\cos 40 \cdot \cos 50} \right)}} \cdot \hfill \\
\hfill \\
\frac{1}{{{2^2}\left( {\cos 48 \cdot \cos 44} \right)}} \cdot \frac{1}{{2\sin 45}} \hfill \\
\end{gathered} \]

Answer 27

nevertheless it is not the right step.
Here is a more useful step:
\[\begin{gathered}
= \frac{1}{{2\cos 88}} \cdot \frac{1}{{2\cos 84}} \cdot \frac{1}{{2\cos 80}} \cdot ... \cdot \cdot \frac{1}{{2\cos 4}} \cdot \frac{1}{{2\sin 45}} = \hfill \\
\hfill \\
= \frac{1}{{2\left( {\cos 92 + \cos 52} \right)}} \cdot \frac{1}{{2\left( {\cos 92 + \cos 44} \right)}} \cdot \frac{1}{{2\left( {\cos 92 + \cos 36} \right)}} \cdot \hfill \\
\hfill \\
\cdot \frac{1}{{2\left( {\cos 92 + \cos 28} \right)}} \cdot \frac{1}{{2\left( {\cos 92 + \cos 20} \right)}} \cdot \frac{1}{{2\left( {\cos 92 + \cos 12} \right)}} \cdot \hfill \\
\hfill \\
\cdot \frac{1}{{2\left( {\cos 92 + \cos 8} \right)}} \cdot \frac{1}{{2\sin 45}} = \hfill \\
\end{gathered} \]

Answer 28

now, we have:
\[\cos 92 = \cos 90\cos 2 - \sin 90\sin 2 = - \sin 2\]

Answer 29

so, we can write:
\[\begin{gathered}
= - \frac{1}{{{2^7}}}\frac{1}{{\left( {\sin 2 + \cos 52} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 44} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 36} \right)}} \cdot \hfill \\
\hfill \\
\cdot \frac{1}{{\left( {\sin 2 + \cos 28} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 20} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 12} \right)}} \cdot \hfill \\
\hfill \\
\cdot \frac{1}{{\left( {\sin 2 + \cos 8} \right)}} \cdot \frac{1}{{2\sin 45}} = \hfill \\
\end{gathered} \]

Answer 30

oops.. there is a typo:
\[\begin{gathered}
= - \frac{1}{{{2^7}}}\frac{1}{{\left( {\sin 2 + \cos 52} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 44} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 36} \right)}} \cdot \hfill \\
\cdot \frac{1}{{\left( {\sin 2 + \cos 28} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 20} \right)}} \cdot \frac{1}{{\left( {\sin 2 + \cos 12} \right)}} \cdot \hfill \\
\cdot \frac{1}{{\left( {\sin 2 + \cos 4} \right)}} \cdot \frac{1}{{2\sin 45}} \hfill \\
\end{gathered} \]

Answer 31

Sir how will we solve this now?

Answer 32

I tried to solve it this morning, nevertheless I have not any idea :(

Answer 33

Sir the options given are
1).cos1°
2).2^-1/2
3).2^1/2
4).None of these

Answer 34

ok! I will try to continue to search, and Tomorrow I will give the answer, I hope

Answer 35

Thanks sir ....

Answer 36

:)

Answer 37

here is what I got:
\[\huge \frac{{\sqrt 2 }}{{{2^{23}}}}\prod\limits_{n = 1}^{22} {\frac{1}{{\sin \left( {4n - 2} \right)}}} = \frac{1}{{\sqrt 2 }}\]
wherein I made the last step by means of my calculator
@samigupta8

Answer 38

Yes sir that'z ryt....

Answer 39

Sir without the use of calC can't we get it..?

Answer 40

yes! I think so! nevertheless I haven't been able to get the final expression