Question

Please Help:
If cot x = 2 / 3 and x is in quadrant 4, find:
sin(x / 2)

Answer 1

cot^2 + 1 = csc^2; or rather 1/sin^2

Answer 2

\[\frac{1}{cot^2+1}=sin^2\]
\[\sqrt{\frac{1}{cot^2+1}}=sin\]

Answer 3

How did you get sin^2 and cot^2?

Answer 4

from the identity;

cos^2 + sin^2 = 1


cos^2 + sin^2 = 1
---------------- = cot^2 + 1 = 1/sin^2
sin^2

Answer 5

prolly a few too many = signs ... but i think its readable nonetheless :)

Answer 6

then again, i always tend to read these things to quick and over look stuff

Answer 7

cot x = 2 / 3

x = arccot(2/3)

sin(arccot(2/3)) = ?

maybe draw a triangle to help out

Answer 8

sin(x) = 3/sqrt(13)

but how to remember to get sin(x/2)

Answer 9

it has to do with the double angle forumla

Answer 10

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?

Answer 11

\[sin(\frac{x}{2})=\sqrt{\frac{1-\frac{2}{\sqrt{13}}}{2}}\]

\[sin(\frac{x}{2})=\sqrt{\frac{\sqrt{13}-{2}{}}{2\sqrt{13}}}\]

that looks awful :/

Answer 12

sin(x/2)= .4719 if im lucky lol not accounting for any adjustment of signs

Answer 13

http://www.wolframalpha.com/input/?i=sin%28%28arccot%282%2F3%29%29%2F2%29

the wolf agrees

Answer 14

for cos(x/2) I would have to do the same thing?

Answer 15

lol @ the wolf agrees

Answer 16

you could; or if its the same set up as this problem just use the pythag identity:

sin^2 + cos^2 = 1

Answer 17

Thank you :)

Answer 18

yw