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103 Trigonometry Problems

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204 <strong>103</strong> <strong>Trigonometry</strong> <strong>Problems</strong><br />

Lagrange’s Interpolation Formula<br />

Let x 0 ,x 1 ,...,x n be distinct real numbers, and let y 0 ,y 1 ,...,y n be arbitrary real<br />

numbers. Then there exists a unique polynomial P(x)of degree at most n such that<br />

P(x i ) = y i , i = 0, 1,...,n. This polynomial is given by<br />

P(x) =<br />

n∑<br />

i=0<br />

y i (x − x 0 ) ···(x − x i−1 )(x − x i+1 ) ···(x − x n )<br />

(x i − x 0 ) ···(x i − x i−1 )(x i − x i+1 ) ···(x i − x n ) .<br />

Law of Cosines<br />

In a triangle ABC,<br />

|CA| 2 =|AB| 2 +|BC| 2 − 2|AB|·|BC| cos ̸<br />

ABC,<br />

and analogous equations hold for |AB| 2 and |BC| 2 .<br />

Median formula<br />

This is also called the length of the median formula. Let AM be a median in triangle<br />

ABC. Then<br />

|AM| 2 = 2|AB|2 + 2|AC| 2 −|BC| 2<br />

.<br />

4<br />

Minimal Polynomial<br />

We call a polynomial p(x) with integer coefficients irreducible if p(x) cannot be<br />

written as a product of two polynomials with integer coefficients neither of which<br />

is a constant. Suppose that the number α is a root of a polynomial q(x) with integer<br />

coefficients.Among all polynomials with integer coefficients with leading coefficient<br />

1 (i.e., monic polynomials with integer coefficients) that have α as a root, there is<br />

one of smallest degree. This polynomial is the minimal polynomial of α. Let p(x)<br />

denote this polynomial. Then p(x) is irreducible, and for any other polynomial q(x)<br />

with integer coefficients such that q(α) = 0, the polynomial p(x) divides q(x); that<br />

is, q(x) = p(x)h(x) for some polynomial h(x) with integer coefficients.<br />

Orthocenter of a Triangle<br />

The point of intersection of the altitudes.

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