Absolutely clueless, any help would be appreciated, especially if it can be understood by a grade $12$ student (me).
The answer is in the form $2^x$, and we are supposed to find value of $x$.
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*not a homework question, this is a question that had come in one of our exams a few weeks back – Dudeness May 22 '20 at 17:08
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6Hint: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$. – Integrand May 22 '20 at 17:08
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Also, I'm assuming you want to evaluate the trig functions in degrees, but it's probably best to explicitly say so. – Integrand May 22 '20 at 17:09
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I just tried that, we get another series sin1sin3sin5*....in the denominator, how do i progress further? – Dudeness May 22 '20 at 17:10
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@Integrand , yes it is in degrees – Dudeness May 22 '20 at 17:11
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By double-angle, we have $$\require{cancel} \frac{\prod\limits_{n=1}^{45} \cos(2n-1)}{\prod\limits_{n=1}^{45} \sin(4n-2)} = \frac{\cancel{\prod\limits_{n=1}^{45} \cos(2n-1)}}{2^{45}\prod\limits_{n=1}^{45} \sin(2n-1)\cancel{\cos(2n-1)}} =\left(2^{45}\prod_{n=1}^{45}\sin(2n-1)\right)^{-1} $$We can double the product and take the square root $$ \prod_{n=1}^{45}\sin(2n-1) = \sqrt{\prod_{n=1}^{90}\sin(2n-1)} $$This problem was solved by lhf in this question using a comment from Hans Lundmark in this question. The links are excellent reading, I would encourage you to look at them. However, to spoil the result, the second product evaluates to $2^{-89}$, whence we have $$ \left(2^{45}\prod_{n=1}^{45}\sin(2n-1)\right)^{-1} = \left(2^{45}\cdot 2^{-89/2}\right)^{-1}=1/\sqrt{2} $$
Integrand
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