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2002 Winter Diet, paper 2, Q5 - HSN forum

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2002 Winter Diet, paper 2, Q5


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#1 x_Jo_x

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Posted 02 May 2007 - 08:07 PM

i cannot seem to even start this question off and it is really bugging me. I know i have to put in a double angle formula somewhere but i just don't know where or when!

Any help would be great! thank you

#2 Pete

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Posted 02 May 2007 - 08:59 PM

Well, if you look at your formulae list, you'll see that cos2x = 1 - 2sin^2x, so:

\\cos2x - 2sin^2x = 0 \\1 - 2sin^2x - 2sin^2x = 0 \\1 - 4sin^2x = 0 \\4sin^2x = 1 \\sin^2x = \frac{1}{4}

It is relatively easy to solve from there, but post if you have any problems smile.gif  Just remember that it's for 0 < x < 2\pi, so you should end up with four different values.

#3 Indie_Cindy

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Posted 02 May 2007 - 09:06 PM

Hey, im not 100% sure if i did this right, but i got the same answer as thepast paper book.

cos2x - 2sin²x = 0
cos²x - sin²x - 2sin²x = 0
cos²x = sin²x - 2sin²x
cos²x = sinx²(1-2)
cos²x = 1 over 2
cos2x = 1 over 2
cos2x = 60°
cos2x = pie over 6, 5pie over 3
cosx = pie over 6, 5 pie over 6, 7 pie over 6, 11pie over 6

^^^hoped that helped a wee bit smile.gif


#4 x_Jo_x

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Posted 03 May 2007 - 09:18 PM

hey, thanks guys,, that has really helped. The help was very much appreciated! smile.gif





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