# Trig proofWatch

This discussion is closed.
#1
Hi!!!

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

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

Thanks
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#2
Ohh - also is it possible to find the values of x for the equation:

4cot2x = 0

I thought it was imposssible, but apparently you can get 2 ans
0
14 years ago
#3
cansomeone do this too please? (part b)

i have attached the answer, and how do you factorize it like the way its done?
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14 years ago
#4
(Original post by Silly Sally)
Ohh - also is it possible to find the values of x for the equation:

4cot2x = 0

I thought it was imposssible, but apparently you can get 2 ans
It is possible - cotx is defined as tan(90-x) ie tan of complimentary angle. 1/tanx is just a more convenient way of thinking about it, though as tanx tends to infinity, 1/tanx tends to 0.
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#5
(Original post by TheWolf)
Which part? Or is it both?
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#6
(Original post by Bezza)
It is possible - cotx is defined as tan(90-x) ie tan of complimentary angle. 1/tanx is just a more convenient way of thinking about it, though as tanx tends to infinity, 1/tanx tends to 0.
So basically i do

4 tan 2(pi/2 - x) = 0 ???
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14 years ago
#7
(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 = 4[(1-tan^2 x)/(2tanx)]*1/(2sinxcosx)
= (1-tan^2 x)/(sin^2 x) = cosec^2 x - sec^2 x
then use sec^2 x = 1 + tan^2 x, cosec^2 x = 1 + cot^2 x
to get 4cot2xcosec2x = (cotx)^2 - (tanx)^2
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14 years ago
#8
(Original post by Silly Sally)
So basically i do

4 tan 2(pi/2 - x) = 0 ???
Well cotx = 0 at x =+ or - 90, 270 etc so cot2x = 0 at 2x = +- 90, 270 etc, x = +- 45, 135 etc
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#9
(Original post by Bezza)
4cot2xcosec2x = 4[(1-tan^2 x)/(2tanx)]*1/(2sinxcosx)
= (1-tan^2 x)/(sin^2 x) = cosec^2 x - sec^2 x
then use sec^2 x = 1 + tan^2 x, cosec^2 x = 1 + cot^2 x
to get 4cot2xcosec2x = (cotx)^2 - (tanx)^2
Ahhh!!! Thanks!!! I understand now!!! I was going along those lines when i was trying to do this question, but i made a silly mistake and that made me confused
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14 years ago
#10
(Original post by Silly Sally)
Which part? Or is it both?
just b
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#11
(Original post by TheWolf)
just b
Okley dokely!!! I did this question tody i think, so it should come to me!!! It is from Jan 2002 isn't it?
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14 years ago
#12
(Original post by TheWolf)
cansomeone do this too please? (part b)

i have attached the answer, and how do you factorize it like the way its done?
2sin2t(1-cos2t) - 1*(1-cos2t) they've just taken out the common factor of (1-cos2t) so it equals (1-cos2t)(2sin2t-1)
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14 years ago
#13
(Original post by Bezza)
4cot2xcosec2x = 4[(1-tan^2 x)/(2tanx)]*1/(2sinxcosx)
can you explain first line?
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14 years ago
#14
(Original post by Silly Sally)
Ahhh!!! Thanks!!! I understand now!!! I was going along those lines when i was trying to do this question, but i made a silly mistake and that made me confused
you're not called silly sally for nothing
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#15
so you are stuck going from:

2sin2x (1-cos2x) - (1-cos2x) = 0 ----> (1-cos2x)(2sin2x -1) = 0

Well if you look carefully, you should see that the term (1-cos2x) is common in both parts of the function, so you can factorsise this out. Does what i say make sense at all?
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#16
(Original post by Bezza)
you're not called silly sally for nothing
- You now know the real reason behind my name!!!
0
14 years ago
#17
(Original post by TheWolf)
can you explain first line?
cot2x = 1/tan2x, tan2x = 2tanx/(1-tan^2 x) so cot2x = (1-tan^2 x)/2tanx
Similarly, cosec2x = 1/sin2x = 1/(2sinxcosx)
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14 years ago
#18
(Original post by Bezza)
2sin2t(1-cos2t) - 1*(1-cos2t) they've just taken out the common factor of (1-cos2t) so it equals (1-cos2t)(2sin2t-1)
oh yea ofcourse...*tired*
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#19
Are you two worried about P2 - at first i wasn't, but now....
0
14 years ago
#20
(Original post by Bezza)
tan2x = 2tanx/(1-tan^2 x)
ahh where do you get this from?its probs really obvious
0
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