vis at
( tanx - sinx ) / sin'x = 1 / (cosx + cos*x)
ettersom jeg ikke vet hvordan man taster at sinx er opphøyd i 3 osv.. har jeg brukt tegnene ' og *. Disse står for:
' = 3
* = 2
formelen : cos*v + sin*v = 1, hvor *=2 er vel nødvendig for å løse denne oppgaven.
PLZ help me!!
3mx:Trigonometri oppgave(oppgave 244 i boka 3MX matematikk)
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]\frac {tan(x)-sin(x)}{sin^3x} = \frac {1}{cos(x)+cos^2(x)[/tex]
[tex]tan(x)*cos(x) + tan(x)*cos^2x - cos(x)*sin(x) - sin(x)*cos^2(x) = sin^3(x)[/tex]
Vet at tan(x) = sin(x)/cos(x)
[tex]sin(x) + sin(x)cos(x) - sin(x)*cos(x) - sin(x)*cos^2(x) = sin^3(x)[/tex]
[tex]sin(x) = sin^3(x) + sin(x)*cos^2(x)[/tex]
Så vet vi at [tex]cos^2(x) = 1-sin^2(x)[/tex]
[tex]sin(x) = sin^3(x) + sin(x)*(1-sin^2(x)) = sin^3(x) + sin(x) - sin^3(x)[/tex]
[tex]sin(x) = sin(x)[/tex]
Q.E.D
[tex]tan(x)*cos(x) + tan(x)*cos^2x - cos(x)*sin(x) - sin(x)*cos^2(x) = sin^3(x)[/tex]
Vet at tan(x) = sin(x)/cos(x)
[tex]sin(x) + sin(x)cos(x) - sin(x)*cos(x) - sin(x)*cos^2(x) = sin^3(x)[/tex]
[tex]sin(x) = sin^3(x) + sin(x)*cos^2(x)[/tex]
Så vet vi at [tex]cos^2(x) = 1-sin^2(x)[/tex]
[tex]sin(x) = sin^3(x) + sin(x)*(1-sin^2(x)) = sin^3(x) + sin(x) - sin^3(x)[/tex]
[tex]sin(x) = sin(x)[/tex]
Q.E.D
Eller:
[tex]\frac{tan(x)-sin(x)}{sin^3(x)} = \frac{1}{sin^2(x)}(sec(x) - 1) \\ = \frac{1}{1 - cos^2(x)} \ * \ \frac{1-cos(x)}{cos(x)} = \frac{1 - cos(x)}{(1 - cos(x))(1 + cos(x))cos(x)}\\ = \frac{1}{cos(x) + cos^2(x)}[/tex]
[tex]\frac{tan(x)-sin(x)}{sin^3(x)} = \frac{1}{sin^2(x)}(sec(x) - 1) \\ = \frac{1}{1 - cos^2(x)} \ * \ \frac{1-cos(x)}{cos(x)} = \frac{1 - cos(x)}{(1 - cos(x))(1 + cos(x))cos(x)}\\ = \frac{1}{cos(x) + cos^2(x)}[/tex]