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Math Help - Establish the Identity

  1. #1
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    Establish the Identity

    Establish the identity.

    [sin^3 (x) + cos^3 (x)]/[1 - 2 cos^2 (x)] =
    [sec(x) - sin(x)]/[tan(x) - 1]
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by magentarita View Post
    Establish the identity.

    [sin^3 (x) + cos^3 (x)]/[1 - 2 cos^2 (x)] =
    [sec(x) - sin(x)]/[tan(x) - 1]
    Consider the LHS:

    \frac {\sin^3 x + \cos^3 x}{1 - 2 \cos^2 x} = \frac {(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{- \cos 2x}

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


    Now consider the RHS:

    \frac {\sec x - \sin x}{\tan x - 1} = \frac {\frac 1{\cos x} - \sin x}{\frac {\sin x}{\cos x} - 1}

    = \frac {\frac {1 - \sin x \cos x}{\cos x}}{ \frac {\sin x - \cos x}{\cos x}}

    = \frac {1 - \sin x \cos x}{\sin x - \cos x}

    = \frac {1 - \sin x \cos x}{\sin x - \cos x} \cdot \frac {\sin x + \cos x}{\sin x + \cos x}

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

    = LHS

    Thus, LHS \equiv RHS
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  3. #3
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    actually, this is not an identity ... the equation is not true for odd multiples of
    x = \frac{\pi}{2}
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  4. #4
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    Hello, magentarita!

    Establish the identity: . \frac{\sin^3\!x + \cos^3\!x}{1 - 2\cos^2\!x}\;= \;\frac{\sec x - \sin x}{\tan x - 1}
    The left side has a sum of cubes ... factor it.

    . . . \frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{1 - 2 \cos^2\!x}

    . . = \;\frac{(\sin x + \cos x)(\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}} - \sin x\cos x)}{\underbrace{(1 - \cos^2\!x)}_{\text{This is }\sin^2\!x} - \cos^2\!x}

    . . = \;\frac{(\sin x + \cos x)(1 - \sin x\cos x)}{\sin^2\!x - \cos^2\!x}

    . . = \;\frac{{\color{red}\rlap{////////////}}(\sin x + \cos x)(1 - \sin x\cos x)}{{\color{red}\rlap{////////////}}(\sin x + \cos x)(\sin x - \cos x)}

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


    Divide top and bottom by \cos x

    . . \frac{\dfrac{1}{\cos x} - \dfrac{\sin x{\color{red}\rlap{/////}}\cos x}{{\color{red}\rlap{/////}}\cos x}} {\dfrac{\sin x}{\cos x} - \dfrac{{\color{red}\rlap{/////}}\cos x}{{\color{red}\rlap{/////}}\cos x}}

    . . = \;\frac{\sec x - \sin x}{\tan x - 1}

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  5. #5
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    Thanks...

    I thank you all, especially Jhevon and Soroban.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by magentarita View Post
    I thank you all, especially Jhevon and Soroban.
    skeeter had a valid point as well. the secant is undefined for the specified values of x. so that may cause problems. after graphing the two functions, i saw that the graphs were the same and so i overlooked it. i guess the point here was to do the algebraic manipulation to show you know your trig identities and how to use them
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  7. #7
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    Smile No doubt...

    Quote Originally Posted by Jhevon View Post
    skeeter had a valid point as well. the secant is undefined for the specified values of x. so that may cause problems. after graphing the two functions, i saw that the graphs were the same and so i overlooked it. i guess the point here was to do the algebraic manipulation to show you know your trig identities and how to use them
    No doubt but to a struggling math student, the more detail given the better chance for understanding.
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