INFINITE PRODUCT COSINE AND HYPERBOLIC COSINE

Today, we will proof the following relations:



\cos x=\prod_{k=1}^{\infty}\left(1-\frac{4x^2}{\pi^2 (2k-1)^2} \right)(1)


\cosh x=\prod_{k=1}^{\infty}\left(1+\frac{4x^2}{\pi^2 (2k-1)^2} \right)(2)


\tanh\,x=\sum_{k=1}^\infty \frac{8x}{\pi^2(2k-1)^2+4x^2}(3)


\tan\,x=\sum_{k=1}^\infty\frac{8x}{\pi^2(2k-1)^2-4x^2}(4)

Recall the infinite sine product

\sin x = x \prod_{k=1}^{\infty}\left(1-\frac{x^2}{\pi^2 k^2} \right)(5)

\begin{aligned} \cos x &=\frac{2\cos x \sin x}{2 \sin x}\\ &=\frac{\sin 2x}{2 \sin x}\\ &=\frac{2x \prod_{k=1}^{\infty}\left(1-\frac{(2x)^2}{\pi^2 k^2} \right)}{2x \prod_{k=1}^{\infty}\left(1-\frac{x^2}{\pi^2 k^2} \right)} \qquad \qquad \text{(by (5))}\\ &=\frac{2x \prod_{k=1}^{\infty}\left(1-\frac{(2x)^2}{\pi^2 (2k)^2} \right)\prod_{k=1}^{\infty}\left(1-\frac{(2x)^2}{\pi^2 (2k-1)^2} \right)}{2x \prod_{k=1}^{\infty}\left(1-\frac{x^2}{\pi^2 k^2} \right)} \qquad \text{(split the prodduct on the numerator in odd and even terms)}\\ &=\frac{2x \prod_{k=1}^{\infty}\left(1-\frac{x^2}{\pi^2 k^2} \right)\prod_{k=1}^{\infty}\left(1-\frac{(2x)^2}{\pi^2 (2k-1)^2} \right)}{2x \prod_{k=1}^{\infty}\left(1-\frac{x^2}{\pi^2 k^2} \right)} \\ &=\prod_{k=1}^{\infty}\left(1-\frac{4x^2}{\pi^2 (2k-1)^2} \right)\, \blacksquare \end{aligned}


To proof (2), we recall the relation \cos ix= \cosh x, so letting x \mapsto ix in (1)


\begin{aligned} \cosh x&= \prod_{k=1}^{\infty}\left(1-\frac{4(ix)^2}{\pi^2 (2k-1)^2} \right)\\ &= \prod_{k=1}^{\infty}\left(1+\frac{4x^2}{\pi^2 (2k-1)^2} \right) \end{aligned}

letting   x \mapsto \frac{\pi x}{2}   in (1) we obtain


\begin{aligned} \cos \frac{\pi x}{2}&=\prod_{k=1}^{\infty}\left(1-\frac{4 \left( \frac{\pi x}{2}\right)^2}{\pi^2 (2k-1)^2} \right)\\ &=\prod_{k=1}^{\infty}\left(1-\frac{x^2}{ (2k-1)^2} \right)\\ &=\prod_{k=0}^{\infty}\left(1-\frac{x^2}{ (2k+1)^2} \right) \quad \text{shifting index to start in 0}\\ \end{aligned}


\boxed{\cos \frac{\pi x}{2}=\prod_{k=0}^{\infty}\left(1-\frac{x^2}{ (2k+1)^2} \right)}

Similarly, letting x \mapsto \frac{\pi x}{2} in (2) we obtain


\boxed{\cosh \frac{\pi x}{2}=\prod_{k=0}^{\infty}\left(1+\frac{x^2}{(2k+1)^2} \right)}

Corollary


Taking log in both sides of (2)

\begin{aligned} \log\cosh\,x&=\sum_{k=1}^\infty \log\left(1+\frac{4x^2}{\pi^2(2k-1)^2}\right)\\ \tanh\,x&=\sum_{k=1}^\infty \frac{\frac{8x}{\pi^2(2k-1)^2}}{1+\frac{4x^2}{\pi^2(2k-1)^2}} \qquad \text{differentiating w.r. to x}\\ \end{aligned}


\boxed{\tanh\,x=\sum_{k=1}^\infty \frac{8x}{\pi^2(2k-1)^2+4x^2}}


From the relation   \tanh(x)=-i\tan(ix)


\boxed{\tan\,x=\sum_{k=1}^\infty\frac{8x}{\pi^2(2k-1)^2-4x^2}}

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