Question

Sum of 1/(sin2^(n)a) upto n terms

Answer 1

cot a - cot 2^n a

Answer 2

To find the sum of the series $S_n = \sum_{k=1}^{n} \frac{1}{\sin(2^k a)}$, we use a trigonometric identity that allows for a telescopic sum.

Consider the identity: $\cot x - \cot 2x = \frac{\cos x}{\sin x} - \frac{\cos 2x}{\sin 2x}$

We know that $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = 2 \cos^2 x - 1$.

Substitute these into the expression: $\cot x - \cot 2x = \frac{\cos x}{\sin x} - \frac{2 \cos^2 x - 1}{2 \sin x \cos x}$

To combine these terms, find a common denominator: $\cot x - \cot 2x = \frac{2 \cos^2 x}{2 \sin x \cos x} - \frac{2 \cos^2 x - 1}{2 \sin x \cos x}$ $= \frac{2 \cos^2 x - (2 \cos^2 x - 1)}{2 \sin x \cos x}$ $= \frac{2 \cos^2 x - 2 \cos^2 x + 1}{2 \sin x \cos x}$ $= \frac{1}{2 \sin x \cos x}$ $= \frac{1}{\sin 2x}$

So, the identity is $\frac{1}{\sin 2x} = \cot x - \cot 2x$.

Now, let's apply this identity to the general term of our series, $T_k = \frac{1}{\sin(2^k a)}$.

To match the form $\frac{1}{\sin 2x}$, we set $2x = 2^k a$.

This implies $x = \frac{2^k a}{2} = 2^{k-1} a$.

Substituting this into the identity: $T_k = \frac{1}{\sin(2^k a)} = \cot(2^{k-1} a) - \cot(2^k a)$

Now, we can write out the terms of the sum $S_n$:

For $k=1$: $T_1 = \frac{1}{\sin(2a)} = \cot(2^{1-1} a) - \cot(2^1 a) = \cot(a) - \cot(2a)$

For $k=2$: $T_2 = \frac{1}{\sin(2^2 a)} = \cot(2^{2-1} a) - \cot(2^2 a) = \cot(2a) - \cot(4a)$

For $k=3$: $T_3 = \frac{1}{\sin(2^3 a)} = \cot(2^{3-1} a) - \cot(2^3 a) = \cot(4a) - \cot(8a)$ ... For $k=n$: $T_n = \frac{1}{\sin(2^n a)} = \cot(2^{n-1} a) - \cot(2^n a)$

Now, sum these terms: $S_n = T_1 + T_2 + T_3 + \dots + T_n$ $S_n = (\cot(a) - \cot(2a)) + (\cot(2a) - \cot(4a)) + (\cot(4a) - \cot(8a)) + \dots + (\cot(2^{n-1} a) - \cot(2^n a))$

This is a telescopic series, meaning that intermediate terms cancel out. $S_n = \cot(a) - \cot(2^n a)$

The sum of the series up to n terms is $\cot(a) - \cot(2^n a)$.