👤
a fost răspuns

Se considera functia f,g:R->R; f(x)=([tex]sin^{2}\alpha [/tex])x+1, g(x)=([tex]cos^{2}\alpha [/tex])x-1.
a)Sa se calculeze f(X)+g(x)
b)Daca [tex] \alpha [/tex] =45°,calclati numarul
N=f(1)+f(2)+f(3)+...+f(2013).

Răspuns :

Albatran
f(x) +g(x) =
(sin²α)x+1, g(x)=(cos²α)x-1.=(sin²α+cos²α)x+1-1=1*x=x


b)sin²45°=((√2)/2)²=2/4=1/2

N=f(1)+f(2)+f(3)+...+f(2013). =
(1/2)*1+1+(1/2) *2 +1 +(1/2) *3+1+....+(1/2)*2013+1=
=(1/2) (1+2+3+...+2013) +1+1+...+1 de 2013 ori 1
=(1/2) *2013*2014/2+2013=
2013*1007/2+2013=2013(1007/2+1)=2013*1009/2=1009*2013/2
 


Alte intrebari