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2004 Paper 1 Q10 - HSN forum

# 2004 Paper 1 Q10

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### #1Misbah

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Posted 18 May 2006 - 05:17 PM

2004 q10 paper 1 help please thanks

### #2Nathan

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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

### #3Misbah

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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/ 10

DEA = 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

### #4Dave

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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

### #5Gillz

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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/ 10

DEA = 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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