Question

Let $N = \prod_{r=1}^{45}sin((2r-1)^\circ)$, then $\sqrt{2 \cdot 2^{45}N}$ is equal to

Answer 1

2^{3/4}

Answer 2

The problem asks us to evaluate $\sqrt{2 \cdot 2^{45}N}$ where $N = \prod_{r=1}^{45}sin((2r-1)^\circ)$.

First, let's write out the terms of $N$: $N = sin(1^\circ) \cdot sin(3^\circ) \cdot sin(5^\circ) \cdots sin((2 \cdot 45 - 1)^\circ)$ $N = sin(1^\circ) \cdot sin(3^\circ) \cdot sin(5^\circ) \cdots sin(89^\circ)$

This is a product of sines of odd angles from $1^\circ$ to $89^\circ$. There are $\frac{89-1}{2} + 1 = 44+1 = 45$ terms in the product.

We use the trigonometric identity $sin(90^\circ - x) = cos(x)$. Let's pair the terms in the product $N$: $sin(89^\circ) = sin(90^\circ - 1^\circ) = cos(1^\circ)$ $sin(87^\circ) = sin(90^\circ - 3^\circ) = cos(3^\circ)$ ... $sin(47^\circ) = sin(90^\circ - 43^\circ) = cos(43^\circ)$

The middle term in the product is $sin(45^\circ)$. So, we can rewrite $N$ by grouping terms: $N = (sin(1^\circ) \cdot sin(3^\circ) \cdots sin(43^\circ)) \cdot sin(45^\circ) \cdot (sin(47^\circ) \cdot sin(49^\circ) \cdots sin(89^\circ))$ Substitute the cosine equivalents for the terms $sin(47^\circ)$ through $sin(89^\circ)$: $N = (sin(1^\circ) \cdot sin(3^\circ) \cdots sin(43^\circ)) \cdot sin(45^\circ) \cdot (cos(43^\circ) \cdot cos(41^\circ) \cdots cos(1^\circ))$

Now, rearrange the terms to form pairs of $sin(x)cos(x)$: $N = (sin(1^\circ)cos(1^\circ)) \cdot (sin(3^\circ)cos(3^\circ)) \cdots (sin(43^\circ)cos(43^\circ)) \cdot sin(45^\circ)$

We use the identity $sin(x)cos(x) = \frac{1}{2}sin(2x)$. The number of pairs $(sin(x)cos(x))$ is the number of terms from $1^\circ$ to $43^\circ$, which is $\frac{43-1}{2} + 1 = 22$ pairs.

Substitute the identity into the expression for $N$: $N = \left(\frac{1}{2}sin(2^\circ)\right) \cdot \left(\frac{1}{2}sin(6^\circ)\right) \cdots \left(\frac{1}{2}sin(86^\circ)\right) \cdot sin(45^\circ)$ There are 22 such terms, each with a factor of $\frac{1}{2}$. $N = \left(\frac{1}{2}\right)^{22} \cdot (sin(2^\circ) \cdot sin(6^\circ) \cdots sin(86^\circ)) \cdot sin(45^\circ)$

We know $sin(45^\circ) = \frac{1}{\sqrt{2}}$. $N = \frac{1}{2^{22}} \cdot \frac{1}{\sqrt{2}} \cdot (sin(2^\circ) \cdot sin(6^\circ) \cdots sin(86^\circ))$ $N = \frac{1}{2^{22} \cdot 2^{1/2}} \cdot (sin(2^\circ) \cdot sin(6^\circ) \cdots sin(86^\circ))$ $N = \frac{1}{2^{22.5}} \cdot (sin(2^\circ) \cdot sin(6^\circ) \cdots sin(86^\circ))$

Let's look at the product $P' = sin(2^\circ) \cdot sin(6^\circ) \cdots sin(86^\circ)$. The angles are $2^\circ, 6^\circ, 10^\circ, \ldots, 86^\circ$. These are of the form $(4k-2)^\circ$. There are 22 terms in this product. This product is of the form $\prod_{k=1}^{n} \sin\left(\frac{(2k-1)\pi}{2n}\right) = \frac{1}{2^{n-1}}$. This formula is for angles $\frac{\pi}{2n}, \frac{3\pi}{2n}, \ldots, \frac{(2n-1)\pi}{2n}$. If we set $2n = 90$, then $n=45$. The angles would be $\frac{\pi}{90}, \frac{3\pi}{90}, \ldots, \frac{89\pi}{90}$. In degrees, this is $2^\circ, 6^\circ, \ldots, 178^\circ$. The product $\prod_{k=1}^{45} sin((2k-1) \cdot 2^\circ) = sin(2^\circ)sin(6^\circ)\cdots sin(178^\circ)$. This product can be written as: $(sin(2^\circ)sin(6^\circ)\cdots sin(86^\circ)) \cdot (sin(94^\circ)sin(98^\circ)\cdots sin(178^\circ))$ Since $sin(180^\circ - x) = sin(x)$, we have: $sin(178^\circ) = sin(2^\circ)$ $sin(174^\circ) = sin(6^\circ)$ ... $sin(94^\circ) = sin(86^\circ)$ So, $P' = (sin(2^\circ)sin(6^\circ)\cdots sin(86^\circ))^2$. The formula gives $\prod_{k=1}^{45} sin(\frac{(2k-1)\pi}{90}) = \frac{1}{2^{45-1}} = \frac{1}{2^{44}}$. Therefore, $(sin(2^\circ)sin(6^\circ)\cdots sin(86^\circ))^2 = \frac{1}{2^{44}}$. Taking the square root (since all sines are positive in the first quadrant), we get: $sin(2^\circ)sin(6^\circ)\cdots sin(86^\circ) = \sqrt{\frac{1}{2^{44}}} = \frac{1}{2^{22}}$.

Now substitute this back into the expression for $N$: $N = \frac{1}{2^{22.5}} \cdot \left(\frac{1}{2^{22}}\right)$ $N = \frac{1}{2^{22.5 + 22}} = \frac{1}{2^{44.5}}$

Finally, we need to find the value of $\sqrt{2 \cdot 2^{45}N}$: $\sqrt{2 \cdot 2^{45}N} = \sqrt{2^{1} \cdot 2^{45} \cdot \frac{1}{2^{44.5}}}$ $= \sqrt{2^{1+45-44.5}}$ $= \sqrt{2^{46-44.5}}$ $= \sqrt{2^{1.5}}$ $= \sqrt{2^{3/2}}$ $= (2^{3/2})^{1/2}$ $= 2^{3/4}$

The final answer is $2^{3/4}$.