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Homework Help: math

Posted by Yulieth on Sunday, November 20, 2011 at 8:24pm.

given sinθ=-5/13 and π<θ<3π/2 find

sin2θ
cos( θ-4π/3)
sin(θ/2)

can some1 explain to me how to do these?

• math - Henry, Tuesday, November 22, 2011 at 8:11pm

  sinA = -5/13 = Y/r,
  
  X^2 + Y^2 = r^2,
  X^2 + (-5)^2 = (13)^2,
  X^2 + 25 = 169,
  X^2 = 169 - 25 = 144,
  X = +- 12.
  X = -12 Because it places our resultant
  vector in the required quadrant(Q3).
  (X,Y) = (-12,-5).
  tanA = Y/X = --5/-12 = 0.416666,
  A = 22.62 Deg. ,
  A = 22.62 + 180 = 202.62(Q3).
  
  180 < 202.62 < 270. Q3.
  
  sin(2A) = sin(405.24) = 0.710.
  
  cos(A-4pi/3) = cos(202.62-240) = 0.7946
  
  sin(A/2) = +- sqrt((1-cosA)/2) =
  sqrt((1-cos202.62)/2) = 0.9806
  

• math - Henry, Wednesday, November 23, 2011 at 3:25pm

  Correction:
  
  sinA = -5/13 = Y/r,
  A = 157.38 Deg.
  
  sin(2A) = sin(314.76) = -0.710.
  
  cos(A-4pi/3)=cos(157.38-240)=0.1284
  
  sin(A/2) = +- sqrt((1-cosA)/2) =
  sqrt((1-cos157.38)/2 = 0.9806.
  

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