se sen=1/3 e x pertence ao quadrante I do ciclo trigonometrico, calcule:?

xuxa

a) sen2x

b) cos2x

-rca-pi-

Observe:

Lembrando que : cos² x + sen² x = 1 ,então;

cos² x + (1/3)² = 1

cos² x = 1 - (1/9)

cos² x = ( 9 - 1 )/9

cos² x = 8/9

cos x = ± √(8/9)

cos x = ± ( 2√2 )/3

Como x Є ao 1º quadrante , logo;

cos x = ( 2√2 )/3

a) sen 2x =

2.sen x .cos x =

2. (1/3) . ( 2√2 )/3 =

R =====> ( 4√2 )/9

b) cos 2x =

cos² x - sen² x =

[ ( 2√2 )/3 ]² - ( 1/3 )² =

( 8/9 ) - ( 1/9 ) =

R ======> 7/9

Good Luck!

Piauí - Teresina , 20/08/2008

Hora 10: 45

Temperatura 30º

Abraços!!!!!!!!!!

Fonte(s):

Minha

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