P2 Nov 03 Past Paper Help! Watch

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rblowers
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#1
Report Thread starter 16 years ago
#1
Just got to my last past paper to do before Monday's exam and have got stuck on my favourite: Trig!

part a)
Theta = x
Prove:
1-tan^2x/1+tan^2x = cos 2x

part b) hence or otherwise prove
tan^2 (pi/8) = 3 -2(root2)

If anybody could help it would be very much appreciated, even better if a link is known to where all the p2 nov 03 solutions are, because it's very difficult to type the algebra!

Good Luck to everyone with their January exams!

Rob
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nikkij1
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#2
Report 16 years ago
#2
(Original post by rblowers)
Just got to my last past paper to do before Monday's exam and have got stuck on my favourite: Trig!

part a)
Theta = x
Prove:
1-tan^2x/1+tan^2x = cos 2x

part b) hence or otherwise prove
tan^2 (pi/8) = 3 -2(root2)

If anybody could help it would be very much appreciated, even better if a link is known to where all the p2 nov 03 solutions are, because it's very difficult to type the algebra!

Good Luck to everyone with their January exams!

Rob
im doing mine 2moro n i havent got the solutions to that papre but have u got a formula sheet with all the trig identities on, try in the book, they hep a lot!!

Good luck !!!
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mikesgt2
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#3
Report 16 years ago
#3
(Original post by rblowers)
Just got to my last past paper to do before Monday's exam and have got stuck on my favourite: Trig!

part a)
Theta = x
Prove:
1-tan^2x/1+tan^2x = cos 2x

part b) hence or otherwise prove
tan^2 (pi/8) = 3 -2(root2)

If anybody could help it would be very much appreciated, even better if a link is known to where all the p2 nov 03 solutions are, because it's very difficult to type the algebra!

Good Luck to everyone with their January exams!

Rob
Here is a hint for part a:

1 + tan^2(x) = sec^2(x)
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rblowers
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Think i've got the first part now, thanks for your help

using 1+ tan^2x = sec^2x

(1-tan^2x)cos^2x

then using tanx = sinx/cosx to expand the brackets

cos^2x - sin^2x = cos2x

There's probably a flaw in there, my trig proofs always tend to be on the dodgy side!

Second part still an utter pain, but cross fingers tommorrows trig question is slightly easier!

Thanks again!

Rob
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mikesgt2
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#5
Report 16 years ago
#5
(Original post by rblowers)
Think i've got the first part now, thanks for your help

using 1+ tan^2x = sec^2x

(1-tan^2x)cos^2x

then using tanx = sinx/cosx to expand the brackets

cos^2x - sin^2x = cos2x

There's probably a flaw in there, my trig proofs always tend to be on the dodgy side!

Second part still an utter pain, but cross fingers tommorrows trig question is slightly easier!

Thanks again!

Rob
Your method for part a is exactly right.

For b, you need to substitute x = pi/8 into the original equation: ( 1 - tan^2 x )/( 1 + tan^2 x ) = cos2x, then rearrange to get the answer.

Good luck for your exam tomorrow, I have P4 at the same time as you.
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