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Thread: [SOLVED] About interval

  1. #1
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    [SOLVED] About interval

    If sin x + cos x = 1/3 and $\displaystyle x\; \epsilon (0, \pi)$ , determine (sin x)^3 - (cos x)^3 !

    $\displaystyle x\; \epsilon (0, \pi)$ means that x lies in interval 0 and $\displaystyle \pi$ including or excluding 0 and $\displaystyle \pi$ ?

    I think it's excluding because if including it must be $\displaystyle x\; \epsilon [0, \pi]$

    Thx
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  2. #2
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    Hello, songoku!

    Is there a typo?
    If the second expression is $\displaystyle \sin^3\!x {\color{red}+} \cos^3\!x$, it's much easier.


    If $\displaystyle \begin{Bmatrix}\;\sin x + \cos x \:=\:\tfrac{1}{3} & {\color{blue}[1]} \;\\ x \in (0, \pi) & {\color{blue}[2]}\; \end{Bmatrix}$, .determine: .$\displaystyle \sin^3\!x {\color{red}+} \cos^3\!x$

    $\displaystyle x \in (0, \pi)$ means that $\displaystyle x$ lies in interval $\displaystyle 0\text{ to }\pi$, excluding $\displaystyle 0\text{ and }\pi$? . . . . Right!

    Square [1]: .$\displaystyle (\sin x + \cos x)^2 \:=\:\left(\tfrac{1}{3}\right)^2 \quad\Rightarrow\quad \sin^2\!x + 2\sin x\cos x + \cos^2\!x \:=\:\tfrac{1}{9} $

    $\displaystyle \text{We have: }\;2\sin x\cos x + \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} \:=\:\tfrac{1}{9} \quad\Rightarrow\quad 2\sin x\cos x \:=\:-\tfrac{8}{9}$

    . . Hence: .$\displaystyle \sin x\cos x \:=\:-\frac{4}{9}$ .[3]


    Cube [1]: .$\displaystyle (\sin x + \cos x)^3 \:=\:\left(\tfrac{1}{3}\right)^3 \quad\Rightarrow\quad \sin^3\!x + 3\sin^2\!x\cos x + 3\sin x\cos^2\!x + \cos^3\!x \:=\:\tfrac{1}{27}$

    $\displaystyle \text{We have: }\;\sin^3\!x + \cos^3\!x + 3\underbrace{\sin x\cos x}_{-\frac{4}{9}}\underbrace{(\sin x + \cos x)}_{\frac{1}{3}} \:=\:\tfrac{1}{27}$

    Therefore: .$\displaystyle \sin^3\!x+\cos^3\!x - \tfrac{4}{9} \:=\:\tfrac{1}{27} \quad\Rightarrow\quad \boxed{\sin^3\!x + \cos^3\!x \;=\;\frac{13}{27}} $

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  3. #3
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    Hi Mr. Soroban

    It's not typo. Finally I've figured out the answer.

    $\displaystyle \sin^3 x-\cos^3 x=(\sin x-\cos x)(\sin^2 x + \sin x \cos x + cos^2 x)$


    To find $\displaystyle \sin x-\cos x$ :

    $\displaystyle (\sin x-\cos x)^2=\sin^2 x - 2\sin x \cos x+cos^2 x$

    $\displaystyle (\sin x-\cos x)^2=1+\frac{8}{9}$

    $\displaystyle (\sin x-\cos x)=\sqrt{\frac{17}{9}}$


    With this, we will get $\displaystyle \sin^3 x-\cos^3 x=\frac{5}{27}\sqrt{9}$

    Thx a lot for your great help Mr. Soroban
    Last edited by mr fantastic; Aug 22nd 2009 at 05:25 AM. Reason: Deleted link to another question.
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