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| Question 716473:  If tanx= 5/3 and secx < 0, then what is the exact value of sinx?
 So far, I have:
 
 Opp: 5, Adj; 3, and Hyp: sqrt(34).  Sin^2x= sinxcosx
 But after that I'm stuck! Please help!
 
 Answer by stanbon(75887)
      (Show Source): 
You can put this solution on YOUR website! If tanx= 5/3 and secx < 0, then what is the exact value of sinx? So far, I have:
 Opp: 5, Adj; 3, and Hyp: sqrt(34). Sin^2x= sinxcosx
 But after that I'm stuck! Please help!
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 Since tan is positive and sec is negative x in an angle in QIII
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 tan = y/x = -5/-3, so y = -5, x = -3
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 Solve for "r": r = sqrt[5^2+3^2] = sqrt(34)
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 Ans: sin(x) = y/r = -5/sqrt(34)
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 Cheers,
 Stan H.
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