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Thalisyane
@Thalisyane
December 2019
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Dado cos x = 1/5, com 0° < x < 90° calcule o seno de x e a tangente de x
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hcsmalves
Sabemos que sen²x + cos²x = 1
sen²x + (1/5)² = 1
sen²x = 1 - 1/25
sen²x = ( 25 - 1)/25
sen²x = 24/25 (como x é do 1° quadrante, temos)
senx = √4.6/25
senx = 2√6/5
tgx = senx/cosx
tgx = 2√6/5 : (1/5)
tgx = 2√6/5 . 5
Tgx = 2√6
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sen²x + (1/5)² = 1
sen²x = 1 - 1/25
sen²x = ( 25 - 1)/25
sen²x = 24/25 (como x é do 1° quadrante, temos)
senx = √4.6/25
senx = 2√6/5
tgx = senx/cosx
tgx = 2√6/5 : (1/5)
tgx = 2√6/5 . 5
Tgx = 2√6