Lurte på om noen vet hvordan man vise denne identiteten?
[tex]\cos(\arcsin x) = sqrt{1-x^2}[/tex]
cos(arcsinx)
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- Ramanujan
- Posts: 260
- Joined: 16/04-2009 21:41
[tex]\small{\text{atm: fys1120, ast1100, mat1120, mat2410 \ . Prev: mat1110, fys-mek1110, mek1100, mat1100, mat-inf1100, inf1100}}[/tex]
Vel, her er en algebraisk tilnærming.
[tex]x = x[/tex]
[tex]x = \sin(\arcsin(x))[/tex]
En identitet man bruker:
[tex]\sin \theta = \sqrt{1-cos^2(\theta)}[/tex]
[tex]x = \sqrt{1 - \cos^2(\arcsin(x))}[/tex]
[tex]x^2 = 1 - \cos^2(\arcsin(x))[/tex]
[tex]\cos^2(\arcsin(x)) = 1 - x^2[/tex]
[tex]\cos(\arcsin(x)) = \sqrt{1 - x^2}[/tex]
Q.E.D
[tex]x = x[/tex]
[tex]x = \sin(\arcsin(x))[/tex]
En identitet man bruker:
[tex]\sin \theta = \sqrt{1-cos^2(\theta)}[/tex]
[tex]x = \sqrt{1 - \cos^2(\arcsin(x))}[/tex]
[tex]x^2 = 1 - \cos^2(\arcsin(x))[/tex]
[tex]\cos^2(\arcsin(x)) = 1 - x^2[/tex]
[tex]\cos(\arcsin(x)) = \sqrt{1 - x^2}[/tex]
Q.E.D
An ant on the move does more than a dozing ox.
Lao Tzu
Lao Tzu
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- Ramanujan
- Posts: 260
- Joined: 16/04-2009 21:41
Tusen takk
Med den metoden klarte jeg utlede de andre identitetene også.

[tex]\small{\text{atm: fys1120, ast1100, mat1120, mat2410 \ . Prev: mat1110, fys-mek1110, mek1100, mat1100, mat-inf1100, inf1100}}[/tex]