Sendo tg(x) = 5/12 e pi < x < 3pi/2, encontre:
a- cos(x)
b- sen(x)
c- cotg(x)
d- sec(x)
e- cossec(x)
a- cos(x)
b- sen(x)
c- cotg(x)
d- sec(x)
e- cossec(x)
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pi<x<3pi/2 ...3ª quadrante ==>sen e cos x<0
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sen(x)/cos(x)=5/12
sen(x)=(5/12)*cos(x)
sen²(x)+cos²(x)=1
[(5/12)*cos(x)]² +cos²(x)=1
(25/144)*cos²(x)+cos²(x)=1
cos²(x) * (25/144+1)=1
cos²(x) * (169/144) =1
cos²(x) = 144/169
cos(x)=12/13 ou cos(x)=-12/13
Como é do 3ª quadrante cos(x)=-12/13
sen(x)=(5/12)*cos(x)
sen(x)=(5/12) * (-12/13) =-5/13
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a)cos(x)=-12/13
b)sen(x)=-5/13
c)cogt(x)=cos(x)/sen(x) =(-12/13)/(-5/13) =12/5
d)sec(x)=1/cos(x) =-13/12
e)cossec(x)=1/sen(x)=1/(-5/13)=-13/5