anonymous
  • anonymous
PLEASE HELP Write cos(t) in terms of csc(t) if the terminal point determined by t is in quadrant II.
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
csc is in inverted sin; and cos is a shifted sin...
amistre64
  • amistre64
cos(t) = sin(t+ (pi/2)) csc(t) = 1/sin(t) sooooo..... cos(t) = 1/csc(t+ (pi/2))
anonymous
  • anonymous
My only options are A. -((csc t)/(sqrt 1 + csc^2t)) B. -((1)/(sqrt 1 + csc^2t)) C. - ((sqrt csc^2t -1)/(csc t)) D. ((1)/(sqrt csc^2t -1)) E. -((1- csc^2t)/(csc t))

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amistre64
  • amistre64
csc is negative in q2; so d is out
amistre64
  • amistre64
we can work it backwards as well; pick an angle you know the cos and csc to in Q2 and we what fits :)
amistre64
  • amistre64
cos(120) = 1/2 ; the sin(120) = -sqrt(3)/2; csc(120) = -2/sqrt(3)
amistre64
  • amistre64
-2 1 --------*k = --- sqrt(3) 2 sqrt(3) k = - -------- 4
amistre64
  • amistre64
i cant get a clear picture in my head of what the problem is asking for yet....
anonymous
  • anonymous
csc(t) = 1/sin(t) \[cos(t) = \sqrt{1-\sin^2t}\] \[\cos(t) = \sqrt{1-1/\csc^2t}\]
anonymous
  • anonymous
It is in the II quadrant, so C is the option!

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