Prove $\prod_{u=0}^{v-1}y\cos\frac{u\pi}v-x\sin\frac{u\pi}v=(-2)^{-v+1}\sum_{w=0}^{\lfloor\frac{v-1}2\rfloor}(-1)^w\binom{v}{2w+1}x^{v-2w-1}y^{2w+1}$

Question

Prove the following trigonometric equation $$\prod_{u=0}^{v-1}y\cos{\frac{u\pi}{v}}-x\sin{\frac{u\pi}{v}}=(-2)^{-v+1}\sum_{w=0}^{\lfloor\frac{v-1}{2}\rfloor}(-1)^{w}\binom{v}{2w+1}x^{v-2w-1}y^{2w+1}$$

The right term is my observation and might not be typed correct. But when the left term is equal to $0$, the image would be $v$ perfectly distributed line like this ($v=9$):

Answer 1

Starting from

$$\prod_{u=0}^{v-1} \left[ y \cos (u\pi/v) - x \sin(u\pi/v) \right]$$

we put $y/\sqrt{x^2+y^2} = \sin\theta$ to get

$$\sqrt{x^2+y^2}^v \prod_{u=0}^{v-1} \sin(\theta-u\pi/v) \\ = \sqrt{x^2+y^2}^v \frac{1}{2^v i^v} \prod_{u=0}^{v-1} [\exp(i(\theta-u\pi/v)) - \exp(-i(\theta-u\pi/v))] \\ = \sqrt{x^2+y^2}^v \frac{1}{2^v i^v} \prod_{u=0}^{v-1} \exp(-i(\theta-u\pi/v)) \prod_{u=0}^{v-1} [\exp(2i(\theta-u\pi/v)) - 1] \\ = \sqrt{x^2+y^2}^v \frac{1}{2^v i^v} \exp(-iv\theta) \exp(+i(v-1)\pi/2) \\ \times \prod_{u=0}^{v-1} [\exp(2i\theta) \exp(-u2\pi i/v)) - 1].$$

Now observe that

$$\prod_{u=0}^{v-1} [z \exp(-u 2\pi i/v)) - 1] \\ = \prod_{u=0}^{v-1} \exp(-u 2\pi i/v) \prod_{u=0}^{v-1} [z - \exp(u 2\pi i/v)] = \exp(-\pi i (v-1)) (z^v-1).$$

Replace $z$ by $\exp(2i\theta)$ to get

$$\sqrt{x^2+y^2}^v \frac{1}{2^v i^v} \exp(-iv\theta) \exp(-i(v-1)\pi/2) (\exp(2iv\theta)-1) \\ = \sqrt{x^2+y^2}^v \frac{(-1)^{v-1}}{2^{v-1}} \frac{\exp(iv\theta)-\exp(-iv\theta)}{2i} \\ = \sqrt{x^2+y^2}^v \frac{(-1)^{v-1}}{2^{v-1}} \sin(v\theta) \\ = \sqrt{x^2+y^2}^v \frac{(-1)^{v-1}}{2^{v-1}} \Im \exp(iv\theta) = \frac{(-1)^{v-1}}{2^{v-1}} \Im \exp(iv\theta) \sqrt{x^2+y^2}^v \\ = \frac{(-1)^{v-1}}{2^{v-1}} \Im \sum_{q=0}^v {v\choose q} (\exp(i\theta)\sqrt{x^2+y^2} -1)^q \\ = \frac{(-1)^{v-1}}{2^{v-1}} \Im \sum_{q=0}^v {v\choose q} (x+iy-1)^q.$$

The coefficient on $y^p$ where for now $0\le p\le v$ is

$$\frac{(-1)^{v-1}}{2^{v-1}} \Im \sum_{q=0}^v {v\choose q} {q\choose p} i^p (x-1)^{q-p}.$$

Observe that

$${v\choose q} {q\choose p} = \frac{v!}{(v-q)! \times p! \times (q-p)!} = {v\choose p} {v-p\choose v-q}$$

and we obtain

$$\frac{(-1)^{v-1}}{2^{v-1}} {v\choose p} \Im \sum_{q=0}^v {v-p\choose v-q} i^p (x-1)^{q-p} \\ = \frac{(-1)^{v-1}}{2^{v-1}} {v\choose p} \Im i^p \sum_{q=p}^v {v-p\choose v-q} (x-1)^{q-p} \\ = \frac{(-1)^{v-1}}{2^{v-1}} {v\choose p} \Im i^p \sum_{q=0}^{v-p} {v-p\choose v-p-q} (x-1)^q \\ = \frac{(-1)^{v-1}}{2^{v-1}} {v\choose p} \Im i^p \sum_{q=0}^{v-p} {v-p\choose q} (x-1)^q \\ = \frac{(-1)^{v-1}}{2^{v-1}} {v\choose p} \Im i^p x^{v-p}.$$

We see from the powers of the imaginary unit that for the imaginary part not to vanish we must have $p=2w+1$ where $0\le 2w+1\le v$ and we get

$$\frac{(-1)^{v-1}}{2^{v-1}} {v\choose 2w+1} (-1)^w x^{v-2w-1}.$$

Summing this over all admissible $w$ will finally yield

$$(-2)^{1-v} \sum_{w=0}^{\lfloor (v-1)/2 \rfloor} {v\choose 2w+1} (-1)^w x^{v-2w-1} y^{2w+1}.$$

Answer 2

To prove the identity

$$\prod_{u=0}^{v-1}y\cos{\frac{u\pi}{v}}-x\sin{\frac{u\pi}{v}}=(-2)^{-v+1}\sum_{w=0}^{\lfloor\frac{v-1}{2}\rfloor}(-1)^{w}\binom{2w+1}{v}x^{v-2w-1}y^{2w+1}$$

you can use the Euler's formula for cosine and sine. $$\cos x = {e^{ix} + e^{-ix} \over 2}, \sin x = {e^{ix} - e^{-ix} \over 2i}$$ You have like starting point

$$\prod_{u=0}^{v-1}y\cos{\frac{u\pi}{v}}-x\sin{\frac{u\pi}{v}}=\prod_{u=0}^{v-1}\left(y\,{e^{i\frac{u\pi}{v}} + e^{-i\frac{u\pi}{v}} \over 2}-x\,{e^{i\frac{u\pi}{v}} - e^{-i\frac{u\pi}{v}} \over 2i}\right)$$