# Trig proof

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15 years ago
#21
(Original post by Silly Sally)
Are you two worried about P2 - at first i wasn't, but now....
Not really - i did it last year and got 92! I have P5 on monday though and then P4 and P6 a week tuesday
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15 years ago
#22
(Original post by TheWolf)
ahh where do you get this from?its probs really obvious
tan 2x = tan (x + x)

Then work from there.
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15 years ago
#23
(Original post by TheWolf)
ahh where do you get this from?its probs really obvious
tan(A+B) = (tanA+tanB)/(1-tanA*tanB) then let A=B=x
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#24
(Original post by TheWolf)
ahh where do you get this from?its probs really obvious
It is one of the double angle formulae.
you can get it if you use tan (A+B) = (tanA + tan B)/(1-(tanAtanB)

EDIT: Damn you all!!! You are all to quick!!!
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15 years ago
#25
(Original post by Bezza)
I have P5 on monday though and then P4 and P6 a week tuesday
Same here!
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#26
(Original post by Bezza)
Not really - i did it last year and got 92! I have P5 on monday though and then P4 and P6 a week tuesday
You are not doing P2 again are you?!?!
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15 years ago
#27
(Original post by Silly Sally)
You are not doing P2 again are you?!?!
No, I didn't realise this thread was only open to P2ers!

Hornblower - which are you most nervous about? I reckon P5 is probably the hardest
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#28
(Original post by Bezza)
No, I didn't realise this thread was only open to P2ers!
Sorry, the way you worded your last post mae me think that he best thing to do is just to ignore me, i am having another "silly sally" moment
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15 years ago
#29
(Original post by Silly Sally)
Hi!!!

Can someone please show me step by step how to prove that:

(cotx)^2 - (tanx)^2 = 4cot2xcosec2x

Thanks
4cot2xcosec2x=4cos2x/(sin^2(2x))
=4(cos^2x-sin^2x)/(4sin^2xcos^2x)
=4cos^2x/4sin^2xcos^2x-4sin^2x/4sin^2xcos^2x
=1/sin^2x-1/cos^2x [1=cos^2x+sin^2x]
=(sin^2x+cos^2x)/sin^2x-(sin^2x+cos^2x)/cos^2x
=1+cot^2x-tan^2x-1
=cot^2x-tan^2

I assume people have already done it but just wanted to do it my self
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