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C2 factorising a trig equation

Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?
Original post by jessyjellytot14
Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?


Simply a mathematical convention. (tan(x))^2 = tan^2 (x).
Original post by jessyjellytot14
Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?


because (tanx)2=(tan2x) (tan x )^2 = (tan^2 x)

if you still don't understand this, you can always say let tan x = y
then 2y2−y−6=0 2y^2 - y - 6 = 0
Reply 3
Avatar for jamestg
jamestg
21
Original post by jessyjellytot14
Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?


To make it much easier and clearer, use tanx=y and so you'd have 2y^2 - y - 6 = 0

I love trig so much, I thought I was going to hate it but now I understand it - it's one of my favourite topics!
Original post by jamestg
To make it much easier and clearer, use tanx=y and so you'd have 2y^2 - y - 6 = 0

I love trig so much, I thought I was going to hate it but now I understand it - it's one of my favourite topics!


Yes, y-substitution might be the better bet here. Simply solve for y, OP, and then solve back for x using x = tan^-1 (y).
Original post by jessyjellytot14
Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?


I think you've misinterpreted the meaning of tan

tan is a function. Not a term itself.

Let's say you have a function of x called x+1. When you square that, that becomes (x+1)2.

Similarly, tan(x) is a function of x. It means we perform something on the x to get a whole new term. In fact, this means that because it's a single term, you could replace tan(x) with u. That would mean you'd get from (2tanx)(tanx) to (2u)(u) which you know equals 2u2 or 2(tan(x))2. Now as an add on, it's just a convention that we write (tan(x))n as tann(x)
(edited 8 years ago)
Reply 6
Avatar for sam_97
sam_97
6
Firstly, it must be noted that:
tan x≠tan×xtan \: x \neq tan \times x so tan2 x≠tan2×xtan^2 \: x \neq tan^2 \times x

In actual fact:
tan2 x=(tan x)2tan^2 \: x = (tan \: x)^2

As Student403 mentioned, it is very important to understand that tan xtan \: x is a function of xx. Writing tantan alone is completely meaningless as it is not a term in itself but a function which must have an input xx in order to have an output yy.

To help you visualise the problem, if you let u=tan x u = tan \: x then your first equation would become 2u2−u=62u^2 - u = 6. If you factorise this and put tan xtan\:x back in, you will arrive at the correct answer.

2u2−u−6=02u^2 - u - 6 = 0

(2u+3)(u−2)=0(2u + 3)(u - 2) = 0

(2tan x+3)(tan x−2)=0(2tan \:x + 3)(tan \:x - 2) = 0

Hope that helps :smile:
(edited 8 years ago)
Original post by jamestg
To make it much easier and clearer, use tanx=y and so you'd have 2y^2 - y - 6 = 0

I love trig so much, I thought I was going to hate it but now I understand it - it's one of my favourite topics!


Original post by LelouchViRuge
because (tanx)2=(tan2x) (tan x )^2 = (tan^2 x)

if you still don't understand this, you can always say let tan x = y
then 2y2−y−6=0 2y^2 - y - 6 = 0


Oh okay thank you, that makes so much more sense now! :smile:

Say if you had 8sin22x - 2sin2x - 3 = 0 , and you let sinx= y, how would you express y?

Because 8sin2x would then be 8y2 but what about 8sin22x ?
(edited 8 years ago)
Reply 8
Avatar for jamestg
jamestg
21
Original post by jessyjellytot14
Oh okay thank you, that makes so much more sense now! :smile:

Say if you had 8sin22x - 2sin2x - 3 = 0 , and you let sinx= y, how would you express y?


4y^2 - y - 3 = 0 I think??? @Zacken

I have only had to factorise when the only extra numbers are coefficients
Reply 9
Avatar for Zacken
Zacken
22
Original post by jessyjellytot14
Oh okay thank you, that makes so much more sense now! :smile:

Say if you had 8sin22x - 2sin2x - 3 = 0 , and you let sinx= y, how would you express y?

Because 8sin2x would then be 8y2 but what about 8sin22x ?


In this case, you'd let y=sin⁡2xy = \sin 2x to get 8y2−2y−3=08y^2 - 2y - 3=0, where y=sin⁡2x≠sin⁡xy = \sin 2x \neq \sin x.

This would then get you a quadratic in yy that you can solve and back-substitute y=sin⁡2xy = \sin \mathbf{2}x.

In general, if you are given αsin⁡2ax+βsin⁡ax+γ\alpha \sin^2 ax + \beta \sin ax + \gamma then make the substitution y=sin⁡axy= \sin ax.
(edited 8 years ago)
Original post by Zacken
In this case, you'd let y=sin⁡2xy = \sin 2x to get 8y2−2y−3=08y^2 - 2y - 3=0, where y=sin⁡2x≠sin⁡xy = \sin 2x \neq \sin x.

This would then get you a quadratic in yy that you can solve and back-substitute y=sin⁡2xy = \sin \mathbf{2}x.

In general, if you are given αsin⁡2ax+βsin⁡ax+γ\alpha \sin^2 ax + \beta \sin ax + \gamma then make the substitution y=sin⁡axy= \sin ax.


Ahhhhhh it makes so much sense, thank you!
Reply 11
Avatar for Zacken
Zacken
22
Original post by jamestg
Ahhhhhh it makes so much sense, thank you!


:biggrin:

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