anonymous
  • anonymous
Please Help: If cot x = 2 / 3 and x is in quadrant 4, find: sin(x / 2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
cot^2 + 1 = csc^2; or rather 1/sin^2
amistre64
  • amistre64
\[\frac{1}{cot^2+1}=sin^2\] \[\sqrt{\frac{1}{cot^2+1}}=sin\]
anonymous
  • anonymous
How did you get sin^2 and cot^2?

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amistre64
  • amistre64
from the identity; cos^2 + sin^2 = 1 cos^2 + sin^2 = 1 ---------------- = cot^2 + 1 = 1/sin^2 sin^2
amistre64
  • amistre64
prolly a few too many = signs ... but i think its readable nonetheless :)
amistre64
  • amistre64
then again, i always tend to read these things to quick and over look stuff
amistre64
  • amistre64
cot x = 2 / 3 x = arccot(2/3) sin(arccot(2/3)) = ? maybe draw a triangle to help out
amistre64
  • amistre64
|dw:1332820207348:dw|
amistre64
  • amistre64
sin(x) = 3/sqrt(13) but how to remember to get sin(x/2)
amistre64
  • amistre64
it has to do with the double angle forumla
amistre64
  • amistre64
cos(2x) = cos^2(2x/2) - sin^2(2x/2) cos(x) = cos^2(x/2) - sin^2(x/2) cos(x) = 1-2sin^2(x/2) 1-cos(x) = 2sin^2(x/2) (1-cos(x))/2 = sin^2(x/2) sqrt(1-cos(x))/2) = sin(x/2) right?
amistre64
  • amistre64
\[sin(\frac{x}{2})=\sqrt{\frac{1-\frac{2}{\sqrt{13}}}{2}}\] \[sin(\frac{x}{2})=\sqrt{\frac{\sqrt{13}-{2}{}}{2\sqrt{13}}}\] that looks awful :/
amistre64
  • amistre64
sin(x/2)= .4719 if im lucky lol not accounting for any adjustment of signs
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sin%28%28arccot%282%2F3%29%29%2F2%29 the wolf agrees
anonymous
  • anonymous
for cos(x/2) I would have to do the same thing?
anonymous
  • anonymous
lol @ the wolf agrees
amistre64
  • amistre64
you could; or if its the same set up as this problem just use the pythag identity: sin^2 + cos^2 = 1
anonymous
  • anonymous
Thank you :)
amistre64
  • amistre64
yw

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