2004 q10 paper 1 help please thanks
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2004 Paper 1 Q10
Started by Misbah, May 18 2006 05:17 PM
4 replies to this topic
#2
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Posted 18 May 2006 - 06:01 PM
CE = 
10 hence sinx = 1/ 10 cosx = 3/ 10
DEA = 2x +90
cosDEA = cos(2x+90)
cos(2x+90) = cos2xcos90 - sin2xsin90
= -sin2x = -2sinxcosx
=-2*(1/ 10) * 6/ 10
= -6 / 10
10 hence sinx = 1/ 10 cosx = 3/ 10DEA = 2x +90
cosDEA = cos(2x+90)
cos(2x+90) = cos2xcos90 - sin2xsin90
= -sin2x = -2sinxcosx
=-2*(1/
10) * 6/ 10= -6 / 10
#3
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Posted 18 May 2006 - 07:25 PM
QUOTE(nathanm @ May 18 2006, 07:01 PM) 
CE =
10 hence cosx = 1/ 10 sinx = 3/ 10DEA = 2x +90
cosDEA = cos(2x+90)
cos(2x+90) = cos2xcos90 - sin2xsin90
hey i got up to there aswell but how do u get the bit aftr tht
#4
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Posted 18 May 2006 - 07:48 PM
use trig identitites for cos2x and replace cos 90 and sin 90 with there exact values
If i am not here i am somewhere else
#5
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Posted 01 January 2007 - 08:01 PM
QUOTE(nathanm @ May 18 2006, 06:01 PM)
CE =
10 hence sinx = 1/ 10 cosx = 3/ 10DEA = 2x +90
cosDEA = cos(2x+90)
cos(2x+90) = cos2xcos90 - sin2xsin90
= -sin2x = -2sinxcosx
=-2x (1/
10) x 6/ 10= -6 / 10
The second last line should be changed to -2 * (1/
10) * (3/ 10)
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