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Math Help - Range of trigonometric expression?

  1. #1
    Super Member fardeen_gen's Avatar
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    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
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  2. #2
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    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.
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  3. #3
    Super Member fardeen_gen's Avatar
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    How is the final expression less than or equal to one? Please explain.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    [(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..

    <br />
\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 x\not=0)

    = \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)}
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  5. #5
    Super Member fardeen_gen's Avatar
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    Is the problem wrong then?
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  6. #6
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    Hello,

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

    i'll put my simplification until some point..

    <br />
\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 x\not=0)

    = \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 \frac{\tan^6x+3\tan^4x-8\tan^2x+2}{\tan^4x(1-\tan^2x)}
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  7. #7
    Super Member fardeen_gen's Avatar
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    - \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}" alt="- \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]
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  8. #8
    Moo
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    Quote Originally Posted by fardeen_gen View Post
    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 -1+\frac{4}{1-\tan^2x}+\frac{8}{\tan^4x}-\frac{6}{\tan^4x(1-\tan^2x)}

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  9. #9
    Super Member fardeen_gen's Avatar
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    - \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.
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  10. #10
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    Quote Originally Posted by fardeen_gen View Post
    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.
    Oh sorry !

    Let's take it from scratch.

    \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 \frac{\tan^4x}{\tan^4x}, we get :

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

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

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

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

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


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