blackops23
Member
- Joined
- Dec 15, 2010
- Messages
- 428
- Gender
- Male
- HSC
- 2011
Show that (i) cos 46 (degrees) ~ [(root 2)/360)(180 - pi)]
(ii) sin 29 (degrees) ~ (1/360)(180(root 3) - 1))
obv it has something to do with 45 + 1 = 46, and 30 - 1 = 29...
i then tried using like...limx--> 0 cosx - 1/x = 1 and lim x-->0 sinx/x = 1....because u no - 1 degree is very close to 0.
so for part (i) = (1/(root 2))(cos 1 - sin 1)....
now cos 1 = cos (pi/180), therefore [cos(pi/180) -1 ]divided by (pi/180) is almost equal to 1.....so i replaced cos 1 with = pi/180 + 1....
and similarly replaced sin 1 with pi/180....and for part (i) I got 1/(root 2)....=( and for part 2 I got (1/360)(180 - (pi)(1- root(3)).???
any help please?
(ii) sin 29 (degrees) ~ (1/360)(180(root 3) - 1))
obv it has something to do with 45 + 1 = 46, and 30 - 1 = 29...
i then tried using like...limx--> 0 cosx - 1/x = 1 and lim x-->0 sinx/x = 1....because u no - 1 degree is very close to 0.
so for part (i) = (1/(root 2))(cos 1 - sin 1)....
now cos 1 = cos (pi/180), therefore [cos(pi/180) -1 ]divided by (pi/180) is almost equal to 1.....so i replaced cos 1 with = pi/180 + 1....
and similarly replaced sin 1 with pi/180....and for part (i) I got 1/(root 2)....=( and for part 2 I got (1/360)(180 - (pi)(1- root(3)).???
any help please?
Last edited: