show that tan3A cotA cannot have values between 1/3 and 3?

Question

1 pre university question

Answer 1

Sounds like a polynomial inequality in t=tanA :

Crunching out the basics...
tan2A = 2tanA/(1-tan²A) = 2t/(1-t²)

tan3A = ( t + 2t/(1-t²) ) / (1 - t*2t/(1-t²))
= ( t(1-t²) + 2t ) / ((1-t²) - 2t²)
= (3t - t³) / (1 - 3t²)

Then
tan3A cotA = (1/t)(3t - t³) / (1 - 3t²)
= (3 - t²) / (1 - 3t²)

So we are asked to show that f(t)= (3-t²) / (1-3t²)
cannot have values between 1/3 and 3.

Noting that f(t) is a continuous fn. It's also an even fn: f(-t)=f(t), i.e. it's symmetric about the axis t=0. (So strictly you only need to consider the t ≥ 0 half)

The general behavior of f(t) has three regions as follows, taking samples values at t=±1, ±2 :
f(0) = 3
f(±1) = -1
Lim_t→±∞ f(t) = 1/3
f(±2) = (3-4)/(1-3*4) = 1/11

The denominator→0 when (1-3t²)=0, i.e. t=±1/√3
So f(t)→∞ as t→±1/√3

So for -∞ < t < -1/√3 and 1/√3 < t < +∞ :
f(t) < 1/3
To prove this, it suffices to note that f(2)<1/3 and f(t) has no extrema on those intervals ( f'(t) != 0 anywhere on those intervals ).

For -1/√3 < t < 1/√3 :
f(t) > 3
(To prove this, you show f(t) has one minimum at 0
(by solving f'(t)=0), where f(0)=3 at that minimum)


NOTE: after doing all this, it seems easier in retrospect to
consider a clever denominator transform instead of f(t)= (3-t²) / (1-3t²)
Set s=(1-3t²)
Then numerator (3-t²) = (8/3 + 1/3-t²) = 8/3 + s/3 = (8+s)/3

Then f(t) is transformed to:
f(s)= (8+s)/3 / s
f(s) = 1/3 (1 + 8/s)

And under the transform s=1-3t², the domain -∞
The behavior of f(s) is pretty obvious , it decreases asymptotically towards a minimum of (1/3+8/3)=3 as s→+1
Since the domain of s never goes >1, then f(s) can never go below 3.
This method is faster because we eliminated the need for differentiation.

Answer 2

smci has the right idea. But the transformation with the s substitution is unnecessary.

f(t) = (3 - t²)/(1 - 3t²) = {(1/3)(9 - 3t²)}/(1 - 3t²)
= {(1/3)(8 + 1 - 3t²)}/(1 - 3t²)
= {(1/3)(8) + (1/3)(1 - 3t²)}/(1 - 3t²)
= 1/3 + (8/3)/(1 - 3t²)

For t element (-1/√3,1/√3) f(t) ≥ 3 (minimum is 3 at t = 0)
For {t < -1/√3} and {t > 1/√3} f(t) ≤ 1/3
f(t) is undefined for {t = ± 1/√3}

Answer 3

tan3AcotA=3-tan^2A/1-3tan^A
let tan3Acot3A=x
after simplyfying it, we have
3-tan^2(A)=x-3tan^2A*x
=>(3x-1)tan^2A- x+3=0
let tanA=t
so we have
(3x-1)t^2-x+3=0 which is a quad. eq.
using shridhacharya formula
-(3x-1)(3-x)>=0 (t is real)
(3x-1)(x-3)>=0
after solving this we will get that the values of x(tan3AcotA) cannot lie between 1/3 and 3

Answer 4

tan3A=sin3A/cos3A=(3sinA-4sin^3A)/(4cos^3A-3cosA)

tan3A*cotA= (3sinA-4sin^3A)/(4cos^3A-3cosA)*(cosA/sinA)

=(3-4sin^2A)/(4cos^2A-3)
equal it to 3
3=(3-4sin^2A)/(4cos^2A-3)
12cos^2A-9=3-4sn^2A

-4sin^2A=4cos^2A-4
12cos^2A-9=3-4+4cos^2A
8cos^2A=8
CosA=+ or -1
cosA is valid between +1 and -1 only

Similarly equal it to (1/3)
then you get
9-12sin^2A=4cos^A-3

9-12sin^2A=1-4sin^2A
8=8sin^2A
sinA=+ or-1

Sin values are real only between 1 and -1
Any value between 1/3 and 3 will give out cosA and sin A values outside the limit of 1/3 and 3

hence tan3AcotA cant have values between 1/3 and 3