# Range of trigonometric expression?

• Jul 18th 2008, 05:30 AM
fardeen_gen
Range of trigonometric expression?
[(3 - tan^2 x)/(tan^2 x)]*[(1 - tan^2 x)/(tan^2 x)]*[(2 tan x)/(1 - tan^2 x)]*[(1 - 3 tan^2 x)/(3 tan x - tan^3 x)] - [3cot^2 x - 1]*[cot^2 x - 1] is the given expression.

Prove this expression is less than or equal to 1.

I couldn't prove it to be less than or equal to 1.(Maybe I messed up in the intermediate steps where after simplification I got the expression = -[(cot^2 x + 1)]^2
• Jul 18th 2008, 05:49 AM
TKHunny
That's quite a mess.

Your final expression is less than or equal to zero (0). Squaring makes positive.

That is either sufficient to the task or you wandered off somwehere.
• Jul 18th 2008, 05:53 AM
fardeen_gen
How is the final expression less than or equal to one? Please explain.
• Jul 18th 2008, 05:56 AM
kalagota
Quote:

Originally Posted by fardeen_gen
[(3 - tan^2 x)/(tan^2 x)]*[(1 - tan^2 x)/(tan^2 x)]*[(2 tan x)/(1 - tan^2 x)]*[(1 - 3 tan^2 x)/(3 tan x - tan^3 x)] - [3cot^2 x - 1]*[cot^2 x - 1] is the given expression.

Prove this expression is less than or equal to 1.

I couldn't prove it to be less than or equal to 1.(Maybe I messed up in the intermediate steps where after simplification I got the expression = -[(cot^2 x + 1)]^2

it cannot be that since that is equal to -csc^4 x, right?

i'll put my simplification until some point..

$\displaystyle \frac{3-\tan^2 x}{\tan^2 x} \cdot \frac{1-\tan^2 x}{\tan^2 x} \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \underbrace{\frac{1-3\tan^2 x}{3\tan -\tan^3 x}}_{\tan x(3-\tan^2 x)} - \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}$ (we assume that $\displaystyle x\not=0$)

$\displaystyle = \frac{2(1-3\tan^2 x)}{(\tan^4 x)(1-\tan^2 x)}-\frac{3-\tan^2x}{1-\tan^2 x} = \frac{2(1-3\tan^2 x)-(\tan^4x)(3-\tan^2x)}{(\tan^4 x)(1-\tan^2 x)}$
• Jul 18th 2008, 08:28 AM
fardeen_gen
Is the problem wrong then?
• Jul 18th 2008, 10:55 AM
Moo
Hello,

Quote:

Originally Posted by kalagota
it cannot be that since that is equal to -csc^4 x, right?

i'll put my simplification until some point..

$\displaystyle \frac{3-\tan^2 x}{\tan^2 x} \cdot \frac{1-\tan^2 x}{\tan^2 x} \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \underbrace{\frac{1-3\tan^2 x}{3\tan -\tan^3 x}}_{\tan x(3-\tan^2 x)} - \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}$ (we assume that $\displaystyle x\not=0$)

$\displaystyle = \frac{2(1-3\tan^2 x)}{(\tan^4 x){\color{red}(1-\tan^2 x)}}-\frac{3-\tan^2x}{1-\tan^2 x} = \dots$

Isn't there a typo here ? (in red) it shouldn't be here.

This gives $\displaystyle \frac{\tan^6x+3\tan^4x-8\tan^2x+2}{\tan^4x(1-\tan^2x)}$
• Jul 18th 2008, 11:02 PM
fardeen_gen
Quote:

$\displaystyle - \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}$

The expression was - [3 cot^2 x -1]. [cot^2 x - 1]
• Jul 18th 2008, 11:09 PM
Moo
Quote:

Originally Posted by fardeen_gen
The expression was - [3 cot^2 x -1]. [cot^2 x - 1]

He just said he multiplied by tanē/tanē, which is possible because it's equal to 1.

For the simplification, I get to $\displaystyle -1+\frac{4}{1-\tan^2x}+\frac{8}{\tan^4x}-\frac{6}{\tan^4x(1-\tan^2x)}$

(Worried)
• Jul 18th 2008, 11:15 PM
fardeen_gen
Quote:

$\displaystyle - \frac{3\cot^2 x-1}{\cot^2 x-1}$
I could understand multiplying by tan^2 x/tan^2 x. But the part of the expression was not a fraction. It was a product.( ie (3 cot^2 x - 1) multiplied by (cot^2 x - 1) ,not divided by (cot^2 x - 1))

God save us! It seems the problem is jinxed. The expression simplify refuses to behave and simply into a reasonable quantity.(Thinking)
• Jul 18th 2008, 11:47 PM
Moo
Quote:

Originally Posted by fardeen_gen
I could understand multiplying by tan^2 x/tan^2 x. But the part of the expression was not a fraction. It was a product.( ie (3 cot^2 x - 1) multiplied by (cot^2 x - 1) ,not divided by (cot^2 x - 1))

God save us! It seems the problem is jinxed. The expression simplify refuses to behave and simply into a reasonable quantity.(Thinking)

Oh sorry !

Let's take it from scratch.

$\displaystyle \frac{3-\tan^2 x}{\tan^2 x} \cdot \frac{1-\tan^2 x}{\tan^2 x} \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \underbrace{\frac{1-3\tan^2 x}{3\tan -\tan^3 x}}_{\tan x(3-\tan^2 x)} - (3\cot^2 x-1)(\cot^2 x-1)$

Simplifying the left term and multiplying the right term by $\displaystyle \frac{\tan^4x}{\tan^4x}$, we get :

$\displaystyle \frac{2(1-3\tan^2x)}{\tan^4x}-\frac{[\tan^2x (3 \cot^2x-1)][\tan^2x(\cot^2x-1)]}{\tan^4x}$

$\displaystyle =\frac{2(1-3 \tan^2x)-(3-\tan^2x)(1-\tan^2x)}{\tan^4x}$

$\displaystyle =\frac{2-6\tan^2x-3+4 \tan^2x-\tan^4x}{\tan^4x}$

$\displaystyle =-\frac{\tan^4x+2 \tan^2x+1}{\tan^4x}$

$\displaystyle =\boxed{-\frac{(\tan^2x+1)^2}{\tan^4x}}$

And this is always less than 1 because it's negative. This is only true if $\displaystyle x \neq 0 \bmod \pi$
So you will have to find the value if x=0.