Olimpíadas ⇒ (Alemanha) Intervalo de valores Tópico resolvido

[undefinied3]

Dada a função
[tex3]F(x)=sen(3x)sen^3(x)+cos(3x)cos^3(x)-\frac{3}{4}cos(2x)[/tex3]
Prove que [tex3]-\frac{1}{4} \leq F(x) \leq \frac{1}{4}[/tex3]
Para todo x real.

[undefinied3]

[tex3]sen(3x)sen^3(x)+cos(3x)cos^3(x)-\frac{3}{4}cos(2x)=(-4sen ^3(x)+3sen(x))sen^3(x)+(4cos^3(x)-3cos(x))cos^3(x)-\frac{3}{4}(cos^2(x)-sen^2(x))[/tex3]
[tex3]-4sen^6(x)+3sen^4(x)+4cos^6(x)-3cos^4(x)-\frac{3}{4}(cos^2(x)-sen^2(x))[/tex3]
[tex3]4(cos^6(x)-sen^6(x))-3(cos^4(x)-sen^4(x))-\frac{3}{4}(cos^2(x)-sen^2(x))[/tex3]
[tex3]cos^6(x)-sen^6(x)=(cos^2(x)-sen^2(x))(cos^4(x)+cos^2(x)sen^2(x)+sen^4(x))[/tex3]
[tex3]cos^4(x)+cos^2(x)sen^2(x)+sen^4(x)=(cos^2(x)+sen^2(x))^2-sen^2(x)cos^2(x)=\frac{4-sen^2(2x)}{4}[/tex3]
[tex3]cos^4(x)-sen^4(x)=(cos^2(x)-sen^2(x))(cos^2(x)+sen^2(x))=cos(2x)[/tex3]
[tex3]4cos(2x)(\frac{4-sen^2(2x)}{4})-3cos(2x)-\frac{3}{4}cos(2x)=[/tex3]
[tex3]cos(2x)(4-sen^2(2x))-cos(2x)(3+\frac{3}{4})=[/tex3]
[tex3]cos(2x)(4-sen^2(2x)-3-\frac{3}{4})=[/tex3]
[tex3]cos(2x)(cos(2x)^2-\frac{3}{4})=[/tex3]
[tex3]\frac{4cos^3(2x)-3cos(2x)}{4}=[/tex3]
[tex3]\frac{cos(6x)}{4}[/tex3]
[tex3]\therefore -\frac{1}{4} \leq F(x) \leq \frac{1}{4}[/tex3]