Starting from sin2x=2x product from k=1 to infinity (1
Starting from $\displaystyle\sin2x=2x\prod_{k=1}^{\infty}\left(1-\frac{4x^2}{k^2\pi^2}\right)$, show that $\displaystyle\cos x=\prod_{k=1}^{\infty}\left[1-\frac{4x^2}{(2k-1)^2\pi^2}\right]$.
Explanation of Method
Dividing the given product for $\sin2x=2\sin x\cos x$ by Euler's product for $\sin x$ isolates $\cos x$ as a ratio of two products. Splitting the numerator's index $k$ into even ($k=2j$) and odd ($k=2j-1$) parts shows the even part exactly matches (and cancels against) the denominator, leaving only the odd part.
- Divide $\sin2x=2\sin x\cos x$ by $\sin x=x\prod(1-x^2/(k^2\pi^2))$ to isolate $\cos x$.
- Split the numerator's product into even- and odd-$k$ parts.
- Cancel the even part against the denominator, leaving the result.
Step 1 · Isolate $\cos x$
Using $\sin2x=2\sin x\cos x$ and Euler's product $\sin x=x\prod_{k=1}^\infty(1-x^2/(k^2\pi^2))$: $$ \cos x = \frac{\sin x\cos x}{\sin x} = \frac{\prod_{k=1}^\infty\left(1-\dfrac{4x^2}{k^2\pi^2}\right)}{\prod_{k=1}^\infty\left(1-\dfrac{x^2}{k^2\pi^2}\right)}. $$
Step 2 · Split even and odd $k$ in the numerator
Writing even $k=2j$ separately from odd $k=2j-1$: $$ \prod_{k=1}^\infty\left(1-\frac{4x^2}{k^2\pi^2}\right) = \underbrace{\prod_{j=1}^\infty\left(1-\frac{4x^2}{(2j)^2\pi^2}\right)}_{=\ \prod_j\left(1-\frac{x^2}{j^2\pi^2}\right)} \cdot \prod_{j=1}^\infty\left(1-\frac{4x^2}{(2j-1)^2\pi^2}\right). $$
Step 3 · Cancel and conclude
The even part exactly equals the denominator's product (same expression, index $j$ instead of $k$), so it cancels: $$ \boxed{\cos x = \prod_{k=1}^{\infty}\left[1-\frac{4x^2}{(2k-1)^2\pi^2}\right]} $$ This matches the Weierstrass factorization directly, since the zeros of $\cos x$ are exactly at $x=\pm(2k-1)\pi/2$. (Numerical check at $x=0.7$: partial product to $10^5$ terms gives $\approx0.764843$, matching $\cos(0.7)\approx0.764842$.)
Summary Table
| Function | Infinite product |
|---|---|
| $\sin x$ | $x\prod_{k\ge1}\left(1-\dfrac{x^2}{k^2\pi^2}\right)$ |
| $\cos x$ | $\prod_{k\ge1}\left(1-\dfrac{4x^2}{(2k-1)^2\pi^2}\right)$ |
Additional Notes
- The even/odd splitting trick used here is the exact converse of the telescoping seen throughout the infinite-product batch (Problems 360–370): there indices were merged via factorization; here a single product is separated by index parity, then one half is recognized and cancelled.
- This product manifestly displays the zeros of $\cos x$ at $x=\pm\pi/2,\pm3\pi/2,\pm5\pi/2,\ldots$ — exactly the classical zero set — providing an independent confirmation via the Weierstrass factorization theorem (Method 3 in the source), which builds the product directly from the known zero locations rather than by algebraic division.
- Problems 395 and 396 obtain the hyperbolic analogues $\sinh x$ and $\cosh x$ from this same family by the substitution $x\to ix$, which flips every minus sign in these products to a plus sign.