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Math Help - Trigonometric Identity Equation

  1. #1
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    Trigonometric Identity Equation [Solved]

    Hello!

    Firstly, thank you for looking at my question! I am new here and although I seem to be able to pick up mathematical concepts rather well, this one problem is absolutely impossible for me to figure out. I know the answer - I just don't know how to get there! I created an image of my problem so its easier to see what I am talking about. Here is the equation:



    Essentially, what I am able to get is this:

    (sin^2 x)/(cos^2 x) = tan^2 x
    -(tan x)/ -(cot x) = 1

    tan^2 x - 1 does not equal tan^2 x

    Any help would be super greatly appreciated, thank you very much!
    -Joshua
    Last edited by vulcangon; November 20th 2009 at 02:54 PM.
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  2. #2
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    working with fractions

    At first glance it looks like you are doing the following:
    <br />
(6-1)/(3-1) = (6/3) - (1/1)<br />
    which isn't valid. Try again, paying attention to the algebra, not the trig.

    This isn't relevant to your problem, but tan / cot doesn't equal 1.
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  3. #3
    Super Member bigwave's Avatar
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    \frac{sin^2x-tanx}{cos^2x-cotx} = tan^2x

    sin^2x-tanx = \frac{sin^2x}{cos^2x}(cos^2x - \frac{1}{tanx})


    sin^2x-tanx  = \frac{sin^2xcos^2x}{cos^2x} - \frac{tan^2x}{tanx}


    sin^2x-tanx = sin^2x-tanx
    Last edited by bigwave; November 20th 2009 at 10:15 AM.
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  4. #4
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    Hello, vulcangon!

    Welcome abaord!


    \frac{\sin^2\!x - \tan x}{\cos^2\!x - \cot x} \:=\:\tan^2\!x

    \frac{\sin^2\!x - \tan x}{\cos^2\!x - \cot x} \;=\;\frac{\sin^2\!x - \dfrac{\sin x}{\cos x}}{\cos^2\!x - \dfrac{\cos x}{\sin x}}

    . . . . . . . . = \;\frac{\sin x\left(\sin x - \dfrac{1}{\cos x}\right)} {\cos x\left(\cos x - \dfrac{1}{\sin x}\right)}

    . . . . . . . . =\; \frac{\sin x\left(\dfrac{\sin x\cos x - 1}{\cos x}\right)} {\cos x\left(\dfrac{\sin x\cos x - 1}{\sin x}\right)}

    . . . . . . . . =\; \frac{\dfrac{\sin x}{\cos x}(\sin x\cos x - 1)}{\dfrac{\cos x}{\sin x}(\sin x\cos x - 1)}

    . . . . . . . . = \;\frac{\tan x}{\cot x}

    . . . . . . . . =\; \frac{\tan x}{\frac{1}{\tan x }}

    . . . . . . . . =\;\tan^2\!x


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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, vulcangon!

    Welcome abaord!



    \frac{\sin^2\!x - \tan x}{\cos^2\!x - \cot x} \;=\;\frac{\sin^2\!x - \dfrac{\sin x}{\cos x}}{\cos^2\!x - \dfrac{\cos x}{\sin x}}

    . . . . . . . . = \;\frac{\sin x\left(\sin x - \dfrac{1}{\cos x}\right)} {\cos x\left(\cos x - \dfrac{1}{\sin x}\right)}

    . . . . . . . . =\; \frac{\sin x\left(\dfrac{\sin x\cos x - 1}{\cos x}\right)} {\cos x\left(\dfrac{\sin x\cos x - 1}{\sin x}\right)}

    . . . . . . . . =\; \frac{\dfrac{\sin x}{\cos x}(\sin x\cos x - 1)}{\dfrac{\cos x}{\sin x}(\sin x\cos x - 1)}

    . . . . . . . . = \;\frac{\tan x}{\cot x}

    . . . . . . . . =\; \frac{\tan x}{\frac{1}{\tan x }}

    . . . . . . . . =\;\tan^2\!x


    Ahhhhhhh Thank you!

    I get it now.

    Also, thanks everyone who responded
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  6. #6
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    But I don't get this so well

    3sin^4 x - 2sin^6 x = 1 - 3cos^4 x + 2cos^6 x
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