Viète's equations are used in some proofs of the Basel problem, which was allegedly solved by Euler.
Viète's equations include the following: given a polynomial,
$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$ with roots $\alpha_1, \alpha_2,...\alpha_n,$
$$a_{n-1}= -a_n\,(\alpha_1+\alpha_2\,+...+\,\alpha_n).$$
This can be applied to a polynomial of the form:
$$b_0 - b_1\, x^2 +b_2\,x^4-\cdots+(-1)^n\,b_n\,x^{2n},$$
in which case we find that,
$$b_0 - b_1\, x^2 +b_2\,x^4-\cdots+(-1)^n\,b_n\,x^{2n}=b_0\left(1-\frac{x^2}{\beta_1^2}\right)\,\left(1-\frac{x^2}{\beta_2^2}\right)\, \cdots\,\left(1-\frac{x^2}{\beta_n^2}\right).$$
And,
$$b_1 = b_0\,\left(\frac{1}{\beta_1^2}\right)+\,\left(\frac{1}{\beta_2^2}\right)\,+\cdots+\,\left(\frac{1}{\beta_n^2}\right).$$
This form allows expressing the second coefficient in:
$$\frac{sin (x)}{x}= 1 - \frac{x^2}{2\cdot 3}+\frac{x^4}{2\cdot 3\cdot 4\cdot 5}-\frac{x^6}{2\cdot3\cdots7}+\cdots$$
as
$$\frac{1}{2\cdot3}= \frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+\cdots.$$
The question is whether Euler truly resorted to Viète's equations, and how big of a mathematical figure Viète name truly is in his own right.