Question

If $\sqrt{2}\sin A=\sin B-{\sin }^{3}B$ and $\sqrt{2}\cos A=\cos B+{\cos }^{3}B$, then the possible value(s) of $\sin (A-B)$ is/are

• A. $\sin (A-B)=-\frac{1}{2\sqrt{2}}\sin 2B$
• B. $\sin (A-B)=\frac{1}{2\sqrt{2}}\sin 2B$
• C. $\sin (A-B)=\frac{1}{3}$
• D. $\sin (A-B)=-\frac{1}{3}$

Answer 1

Answer: (A), (C), (D)

$\sin (A-B)=-\frac{1}{3}$

$\sqrt{2}\sin A=\sin B-{\sin }^{3}B...\left(1\right)$
$\sqrt{2}\cos A=\cos B+{\cos }^{3}B...\left(2\right)$
Now,
$\text{eqn}\left(1\right)\times \cos B-\text{eqn}\left(2\right)\times \sin B,$ gives
$\sqrt{2}(\sin A\cos B-\cos A\sin B)=-{\sin }^{3}B\cos B-{\cos }^{3}B\sin B$
$\Rightarrow \sqrt{2}\sin (A-B)=-\cos B\sin B$
$\Rightarrow \sin (A-B)=\frac{-1}{2\sqrt{2}}\sin 2B...\left(3\right)$

Now,
$2({\cos }^{2}A+{\sin }^{2}A)=(\cos B+{\cos }^{3}B{)}^2+(\sin B-{\sin }^{3}B{)}^2$
$\Rightarrow 2=1+2({\cos }^{4}B-{\sin }^{4}B)+({\cos }^{6}B+{\sin }^{6}B)$
$\Rightarrow 1=2({\cos }^{2}B-{\sin }^{2}B)+({\cos }^{4}B+{\sin }^{4}B-{\cos }^{2}B{\sin }^{2}B)$
$\Rightarrow 1=2\cos 2B+({\cos }^{2}B+{\sin }^{2}B{)}^2-3{\cos }^{2}B{\sin }^{2}B$
$\Rightarrow 1=2\cos 2B+1-\frac{3}{4}{\sin }^{2}2B$
$\Rightarrow 0=8\cos 2B-3{\sin }^{2}2B$
$\Rightarrow 3{\cos }^{2}2B+8\cos 2B-3=0$
$\Rightarrow (3\cos 2B-1)(\cos 2B+3)=0$
$\Rightarrow \cos 2B=\frac{1}{3},\cos 2B\ne -3$
$\sin 2B=\pm \frac{2\sqrt{2}}{3}$
From eqn(3),
$\sin (A-B)=\pm \frac{1}{3}$