Question

2sin2 + 4sin4...180sin180

Answer 1

90cot1°

Answer 2

The given series is $S = 2\sin2^\circ + 4\sin4^\circ + \dots + 180\sin180^\circ$. This can be written as a summation: $S = \sum_{k=1}^{90} (2k)\sin(2k^\circ)$.

We can split the sum into three parts:

1. Terms from $k=1$ to $k=44$: $2\sin2^\circ + 4\sin4^\circ + \dots + 88\sin88^\circ$.
2. The middle term: $90\sin90^\circ$.
3. Terms from $k=46$ to $k=90$: $92\sin92^\circ + 94\sin94^\circ + \dots + 180\sin180^\circ$.

We use the identity $\sin(180^\circ - x) = \sin x^\circ$.

Let's analyze the terms in the third part: $92\sin92^\circ = 92\sin(180^\circ - 88^\circ) = 92\sin88^\circ$. $94\sin94^\circ = 94\sin(180^\circ - 86^\circ) = 94\sin86^\circ$. ... $178\sin178^\circ = 178\sin(180^\circ - 2^\circ) = 178\sin2^\circ$. The last term is $180\sin180^\circ = 180 \times 0 = 0$.

Now, substitute these back into the sum: $S = (2\sin2^\circ + 4\sin4^\circ + \dots + 88\sin88^\circ) + 90\sin90^\circ + (178\sin2^\circ + 176\sin4^\circ + \dots + 92\sin88^\circ) + 0$.

Group the corresponding terms from the first and third parts: $S = (2\sin2^\circ + 178\sin2^\circ) + (4\sin4^\circ + 176\sin4^\circ) + \dots + (88\sin88^\circ + 92\sin88^\circ) + 90\sin90^\circ$.

Each grouped term is of the form $(2k)\sin(2k^\circ) + (180-2k)\sin(2k^\circ)$ for $k=1, 2, \dots, 44$. The sum of coefficients in each pair is $2k + (180-2k) = 180$. So, each pair simplifies to $180\sin(2k^\circ)$.

Therefore, the sum becomes: $S = \sum_{k=1}^{44} 180\sin(2k^\circ) + 90\sin90^\circ$. Since $\sin90^\circ = 1$: $S = 180 \sum_{k=1}^{44} \sin(2k^\circ) + 90$.

Let $P = \sum_{k=1}^{44} \sin(2k^\circ) = \sin2^\circ + \sin4^\circ + \dots + \sin88^\circ$. This is a sum of sines in an arithmetic progression. The first term is $a = 2^\circ$. The common difference is $d = 2^\circ$. The number of terms is $n = 44$.

The formula for the sum of sines in an arithmetic progression is: $\sum_{i=1}^{n} \sin(a + (i-1)d) = \frac{\sin(nd/2)}{\sin(d/2)} \sin(a + (n-1)d/2)$.

Applying the formula to $P$: $P = \frac{\sin(44 \times 2^\circ / 2)}{\sin(2^\circ / 2)} \sin(2^\circ + (44-1)2^\circ/2)$. $P = \frac{\sin(44^\circ)}{\sin(1^\circ)} \sin(2^\circ + 43^\circ)$. $P = \frac{\sin(44^\circ)}{\sin(1^\circ)} \sin(45^\circ)$.

We know that $\sin45^\circ = \frac{1}{\sqrt{2}}$. So, $P = \frac{\sin(44^\circ)}{\sin(1^\circ)} \frac{1}{\sqrt{2}}$.

Now, use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ for $\sin(44^\circ)$: $\sin(44^\circ) = \sin(45^\circ - 1^\circ) = \sin45^\circ\cos1^\circ - \cos45^\circ\sin1^\circ$. $\sin(44^\circ) = \frac{1}{\sqrt{2}}\cos1^\circ - \frac{1}{\sqrt{2}}\sin1^\circ = \frac{1}{\sqrt{2}}(\cos1^\circ - \sin1^\circ)$.

Substitute this back into the expression for $P$: $P = \frac{\frac{1}{\sqrt{2}}(\cos1^\circ - \sin1^\circ)}{\sin1^\circ} \frac{1}{\sqrt{2}}$. $P = \frac{\cos1^\circ - \sin1^\circ}{2\sin1^\circ}$. $P = \frac{1}{2} \left( \frac{\cos1^\circ}{\sin1^\circ} - \frac{\sin1^\circ}{\sin1^\circ} \right)$. $P = \frac{1}{2} (\cot1^\circ - 1)$.

Finally, substitute the value of $P$ back into the expression for $S$: $S = 180 P + 90$. $S = 180 \times \frac{1}{2} (\cot1^\circ - 1) + 90$. $S = 90 (\cot1^\circ - 1) + 90$. $S = 90\cot1^\circ - 90 + 90$. $S = 90\cot1^\circ$.

The final answer is $90\cot1^\circ$.