What Is This ?

Algebra Level pending

Let $P (x)$ be a polynomial of degree $2$ such that

$\begin{cases} P (0) = cos^3 10^\circ \\ P (1) = cos10^\circ sin^2 10^\circ\\ P (2) = 0. \end{cases}$

$\displaystyle P (3) =$ $\displaystyle \frac{\sqrt{a} }{b}$

Find $a^2+b^2$

The answer is 13.

1 solution

Jack D'Aurizio
Apr 16, 2014

Since for a second-degree polynomial the quantity $p(n)-2\cdot p(n+1)+p(n+2)$ is always the same for every $n$ , $P(3) = \cos^3(\pi/18)-3\cos(\pi/18)\sin^2(\pi/18) = \cos(3\cdot\pi/18) = \cos(\pi/6)=\frac{\sqrt{3}}{2}.$

I Solved By Lagrange polynomial

Gabriel Merces - 7 years, 1 month ago

What Is This ?

The answer is 13.

1 solution

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