An interesting Trigonometry problem
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MATHPRITOM
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Prove that, $ sin 1^\circ=cos \frac{1}{2}^\circ .cos \frac{1}{4}^\circ .cos \frac{1}{8}^\circ .cos \frac{1}{16}^\circ ...$
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nafistiham
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Re: An interesting Trigonometry problem
hint:
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Nadim Ul Abrar
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nafistiham
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Re: An interesting Trigonometry problem
a little clarification.....
isnt it a infinite sequence ?
isnt it a infinite sequence ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Nadim Ul Abrar
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Nadim Ul Abrar
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Re: An interesting Trigonometry problem
$sin1=cos\frac{1}{2}.2.sin\frac{1}{2}=cos\frac{1}{2}.cos\frac{1}{4}.sin\frac{1}{4}=....=\frac{cos\frac{1}{2}.cos\frac{1}{4}.cos\frac{1}{8}.cos\frac{1}{16}....sin\frac{1}{2^n}}{\frac{1}{2^n}}$
when $n \to \infty$ then $\frac{1}{2^n} \to 0$
we know when$ \alpha\to 0$ then$ sin\alpha=\alpha $
so $\frac{sin\frac{1}{2^n}}{\frac{1}{2^n}}=1 $hance we done .
when $n \to \infty$ then $\frac{1}{2^n} \to 0$
we know when$ \alpha\to 0$ then$ sin\alpha=\alpha $
so $\frac{sin\frac{1}{2^n}}{\frac{1}{2^n}}=1 $hance we done .
$\frac{1}{0}$
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nafistiham
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Re: An interesting Trigonometry problem
nice oneNadim Ul Abrar wrote:$sin1=cos\frac{1}{2}.2.sin\frac{1}{2}=cos\frac{1}{2}.cos\frac{1}{4}.sin\frac{1}{4}=....=\frac{cos\frac{1}{2}.cos\frac{1}{4}.cos\frac{1}{8}.cos\frac{1}{16}....sin\frac{1}{2^n}}{\frac{1}{2^n}}$
when $n \to \infty$ then $\frac{1}{2^n} \to 0$
we know when$ \alpha\to 0$ then$ sin\alpha=\alpha $
so $\frac{sin\frac{1}{2^n}}{\frac{1}{2^n}}=1 $hance we done .
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: An interesting Trigonometry problem
এইটা হবে তখনই যখন $\alpha$ কে রেডিয়ানে দেয়া হবে। এখানে ডিগ্রীতে দেয়া আছে।Nadim Ul Abrar wrote:
we know when$ \alpha\to 0$ then$ sin\alpha=\alpha $
so $\frac{sin\frac{1}{2^n}}{\frac{1}{2^n}}=1 $hance we done .
অম্লান সাহা
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Nadim Ul Abrar
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Re: An interesting Trigonometry problem
ডিগ্রী (somanupatic) রেডিয়ানamlansaha wrote:এইটা হবে তখনই যখন $\alpha$ কে রেডিয়ানে দেয়া হবে। এখানে ডিগ্রীতে দেয়া আছে।Nadim Ul Abrar wrote:
we know when$ \alpha\to 0$ then$ sin\alpha=\alpha $
so $\frac{sin\frac{1}{2^n}}{\frac{1}{2^n}}=1 $hance we done .
tahole ডিগ্রী te tends to 0 hole রেডিয়ান eo tends to 0 hobe .
:O ...
$\frac{1}{0}$
Re: An interesting Trigonometry problem
$\alpha \to 0$(রেডিয়ানে) হলে $sin\alpha \approx \alpha$ হবে (use calculator)। আমি এটাই বলতে চেয়েছিNadim Ul Abrar wrote:
ডিগ্রী (somanupatic) রেডিয়ান
tahole ডিগ্রী te tends to 0 hole রেডিয়ান eo tends to 0 hobe .
:O ...
অম্লান সাহা