$\sin (A-B) = \pm\frac{1}{3}.$ if....... please help me, how can I approach. I can not understand what to do. How can I start. Help me. If $\sqrt{2}\cos A =\cos B +\cos^3 B$ and $\sqrt{2} \sin A = \sin B - \sin^3 B$ show that $\sin (A-B) =\pm\frac{1}{3}$.
 A: Well, to start with, we can square both equations to get $$2\cos^2A=\cos^2B+2\cos^4B+\cos^6B\\2\sin^2A=\sin^2B-2\sin^4B+\sin^6B$$ Adding these together and doing some factoring yields $$2(\sin^2A+\cos^2A)=\sin^2B+\cos^2B+2(\cos^2B-\sin^2B)(\sin^2B+\cos^2B)+\cos^6B+(\sin^2B)^3,$$ and applying the Pythagorean identity $\sin^2\theta=1-\cos^2\theta$, we find that $$2=1+2(2\cos^2B-1)+\cos^6B+(1-\cos^2B)^3\\2=1+4\cos^2B-2+\cos^6B+1-3\cos^2B+3\cos^4B-\cos^6B\\2=\cos^2B+3\cos^4B$$
Making the substitution $u=\cos^2B$ we have that $$3u^2+u-2=0\\3u^2+3u-2u-2=0\\3u(u+1)-2(u+1)=0\\(3u-2)(u+1)=0$$ so $u=\frac23$ or $u=-1,$ meaning $\cos^2B=-1$ (impossible for real $B$) or $\cos^2B=\frac23.$ Thus, we have $\cos B=\pm\sqrt{\frac23}.$ Let's address these cases separately.

Suppose that $\cos B=\sqrt{\frac23}.$ Applying the Pythagorean identity mentioned above once more, we find that $\sin^B=1-\cos^B=1-\frac23,$ and so $\sin B=\pm\sqrt{\frac13}.$ Then $$\sqrt2\cos A=\sqrt{\frac23}+\frac23\sqrt{\frac23}=\frac53\sqrt{\frac23},$$ and so $$\cos A=\frac53\sqrt{\frac13}.$$ Also, $$\sqrt2\sin A=\pm\sqrt{\frac13}\mp\frac13\sqrt{\frac13}=\pm\frac23\sqrt{\frac13},$$ and so $$\sin A=\pm\frac13\sqrt{\frac23}.$$ (The $\pm$ above corresponds exactly to the $\pm$ of $\sin B.$) Thus, using the identity $$\sin(A-B)=\sin A\cos B-\cos A\sin B,$$ we have $$\sin(A-B)=\left(\pm\frac13\sqrt{\frac23}\right)\left(\sqrt{\frac23}\right)-\left(\frac53\sqrt{\frac13}\right)\left(\pm\sqrt{\frac13}\right)=\pm\frac29\mp\frac59=\mp\frac39=\mp\frac13,$$ as desired.
The $\cos B=-\sqrt{\frac13}$ case is similar.
A: \begin{align}
\sqrt 2 \sin A \cos B = \sin B \cos B \left ( 1-\sin^2B\right ) \tag 1 \\
\sqrt 2 \cos A \sin B = \sin B \cos B \left( 1 + \cos^2 B\right ) \tag 2
\end{align}
Now $(1) - (2)$ and use $\sin(A-B) = \sin A \cos B - \cos A \sin B$
\begin{align}
\sqrt 2  \sin (A-B) &= \sin B \cos B \left( 1-\sin^2B - 1 - \cos^2B\right ) = -\sin B \cos B \\
\sin(A-B) &= -\frac {\sin B\cos B}{\sqrt 2} \tag 3
\end{align}
\begin{align}
\sqrt 2 \cos A \cos B &= \cos^2B \left( 1+\cos^2 B\right ) \tag 4\\
\sqrt 2 \sin A \sin B &= \sin^2B \left( 1-\sin^2B\right ) = \sin^2 B \cos^2B\tag 5
 \end{align}
Now, $(4) + (5)$ and use $\cos(A-B) = \cos A \cos B - \sin A \sin B$
\begin{align}
\sqrt 2 \cos(A-B) &= \cos^2B \left( 1+\cos^2B + \sin^2B\right ) = 2\cos^2B \\
\cos(A-B) &= \sqrt 2 \cos^2B \tag 6
\end{align}
Now use $(3)^2+(6)^2 \equiv 1$, so
$$
\frac {\sin^2B \cos^2B}2 + 2 \cos^4B = 1 \\
\cos^2B \left( \sin^2B + 4 \cos^2B\right ) = 2 \\
\cos^2B \left ( 1 + 3 \cos^2B\right ) = 2 \\
3\cos^4B + \cos^2B - 2 = 0 \\
\cos^2B = \frac {-1 \pm \sqrt{1 + 24}}6 = \frac {-1 \pm 5}6 = \frac 23
$$
Therefore
$$
\sin^2B = 1 - \cos^2B = \frac 13
$$
and
$$
\sin B \cos B = \pm \sqrt{\sin^ 2B \cos^2B} = \pm \frac {\sqrt 2}3
$$
Finally
$$
\sin(A-B) = -\frac 1{\sqrt2} \sin B \cos B = \mp \frac 13
$$
A: As Kaster has identified,
$(6)\displaystyle\sqrt2\cos(A-B)=2\cos^2B=\cos^2B-\sin^2B+1$
$\displaystyle\implies\cos^2B-\sin^2B=\sqrt2\cos(A-B)-1$
$(3)\displaystyle\cos B\sin B=-\sqrt2\sin(A-B)$
We have to eliminate $B$s keeping $A-B$
Using $\displaystyle(\cos^2B-\sin^2B)^2+4\cos^2B\sin^2B=(\cos^2B+\sin^2B)^2=1,$
$\displaystyle\{\sqrt2\cos(A-B)-1\}^2+4\{-\sqrt2\sin(A-B)\}^2=1$
$\displaystyle\implies 2\cos^2(A-B)-2\sqrt2\cos(A-B)+1+8\sin^2(A-B)=1$
Using $\displaystyle\sin^2(A-B)=1-\cos^2(A-B),$
$\displaystyle\implies -6\cos^2(A-B)-2\sqrt2\cos(A-B)+8=0$
Solve the Quadratic Eqaution for $\displaystyle\cos(A-B)$ and find that only one value lies in $\in[-1,1]$
Use $\displaystyle\sin(A-B)=\pm\sqrt{1-\cos^2(A-B)}$
