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Trigonometry proof C3

I have been attempting this question for the past.. 7 hours?
I CAN'T DO IT.
Please help me, someone?

Show how 4tan2x + 3cotxsec2x= 0 becomes 3tan4x - 8tan2x - 3 = 0

I keep ending up with 8tanx-3tanx=0

AAHAHhhhhh noooo
Reply 1
Avatar for davros
davros
16
Original post by VickyImrak
I have been attempting this question for the past.. 7 hours?
I CAN'T DO IT.
Please help me, someone?

Show how 4tan2x + 3cotxsec2x= 0 becomes 3tan4x - 8tan2x - 3 = 0

I keep ending up with 8tanx-3tanx=0

AAHAHhhhhh noooo


That looks a bit of a mess!

Is that really tan 2x or is it meant to be tan2xtan^2 x?

If it's the former, do you know a formula for tan 2x in terms of tan x?
Reply 2
Avatar for VickyImrak
VickyImrak
OP
It is really tan2x
Yep, turns out you just use the formula!
4(2tanx/1-tan2x)+2(1/tanx)(1+tan2x)=0

let tanx = t

8t/(1-t2) + 2/t(1+t2)=0
..
and so on! :smile:
Reply 3
Avatar for ztibor
ztibor
10
Original post by VickyImrak
I have been attempting this question for the past.. 7 hours?
I CAN'T DO IT.
Please help me, someone?

Show how 4tan2x + 3cotxsec2x= 0 becomes 3tan4x - 8tan2x - 3 = 0

I keep ending up with 8tanx-3tanx=0

AAHAHhhhhh noooo


SOme identities
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β→tan⁡2x=2tan⁡x1−tan⁡2x\displaystyle \tan (\alpha + \beta)=\frac{\tan \alpha + \tan \beta}{1-\tan \alpha \cdot \tan \beta} \rightarrow \tan2x=\frac{2\tan x}{1-\tan^2 x}
sec⁡2x=1cos⁡2x=1+tan⁡2x\displaystyle \sec^2 x=\frac{1}{\cos^2 x}=1+\tan^2 x
So the equation
8tan⁡x1−tan⁡2x+3tan⁡x(1+tan⁡2x)=0\displaystyle \frac{8\tan x}{1-\tan^2 x}+\frac{3}{\tan x}(1+\tan^2 x )=0
Multiply by 1−tan⁡2x1-\tan^2 x then by tan⁡x\tan x
8tan⁡2x+3(1−tan⁡4x)=08\tan^2 x +3(1-\tan^4 x)=0
Expand and arrange
(edited 10 years ago)

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