A trigonometric polynomial

Algebra Level 3

If f ( x ) f(x) is a 4-degree polynomial with integer coefficients such that:

f ( sin x ) = f ( cos x ) f(\sin x)=f(\cos x)

f ( 1 ) = 0 f(1)=0

f ( 5 ) = 600 f(5)=-600

find f ( 2 ) f(2) .

1 0 -12 e

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1 solution

s e n 2 x + c o s 2 x = 1 sen^{2} x+cos^{2} x=1

s e n 2 x = 1 c o s 2 x sen^{2} x=1-cos^{2} x and c o s 2 x = 1 s e n 2 x cos^{2} x=1-sen^{2} x

if we multiplicate in cross:

s e n 2 x ( 1 s e n 2 x ) = c o s 2 x ( 1 c o s 2 x ) sen^{2} x(1-sen^{2} x)=cos^{2} x(1-cos^{2} x)

it is the same expression for sen x and for cos x so

f(sen x)=f(cos x) ) \textit{f(sen x)=f(cos x)}) and I we substitute sen x or cos x for x the polynomial appears

f(x) = x 2 ( 1 x 2 ) \textit{f(x)}=x^{2}(1-x^{2})

f(2) = 4 ( 1 4 ) = 12 \textit{f(2)}=4(1-4)=-12

What if f(x)=(x^4)[(1-x^2)^2] or f(x)=2(x^2)(1-x^2)?

X X - 3 years, 2 months ago

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@X X ummm.........well, your given functions don't satisfy f(5)=-600...........

Aaghaz Mahajan - 2 years, 5 months ago

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