# Math Help - [SOLVED] About interval

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

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

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

Thx

2. Hello, songoku!

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

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

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

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

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

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

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

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

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

3. Hi Mr. Soroban

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

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

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

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

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

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

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

Thx a lot for your great help Mr. Soroban