Question

Let $f\left(x\right)=\left|\begin{array}{ccc}\sin x& \sin (x+h)& \sin (x+2h)\\ \sin (x+2h)& \sin x& \sin (x+h)\\ \sin (x+h)& \sin (x+2h)& \sin x\end{array}\right|$.

Answer 1

$\begin{array}{c}f\left(x\right)=\left|\begin{array}{ccc}\sin x& \sin \left(x+h\right)& \sin \left(x+2h\right)\\ \sin \left(x+2h\right)& \sin x& \sin \left(x+h\right)\\ \sin \left(x+h\right)& \sin \left(x+2h\right)& \sin x\end{array}\right|\\ Now,\\ \Delta =si{n}^{3}x+{\sin }^3(x+h)+{\sin }^3(x+2h)-3\sin x\sin (x+h).\\ {\Delta }^1=3{\sin }^2(x+h)\cos (x+h)+6{\sin }^2(x+2h)\cos (x+2h)\\ -3\sin x\cos (x+h)\sin (x+2h)-6\sin x\sin (x+h)\cos (x+2h)\\ Now,LHospitalRule,\\ {\Delta }^{11}=6\sin (x+h){\cos }^2(x+h)-3{\sin }^3(x+h)+24\sin (x+2h){\cos }^2(x+2h)\\ -12{\sin }^3(x+2h)-3\sin x\sin (x+h)\sin (x+2h)-6sinxcos(x+2h)-sinxcos(x+h)cos(x+2h)\\ +12sinxsin(x+h)sin(x+2h).\\ {lim}_{h\to 0}=\frac{\Delta }{h^2}={lim}_{h\to 0}\frac{{\Delta }^{11}}{2}\\ {\Delta }^{11}=6\sin x{\cos }^{2}x-3{\sin }^{3}x+24\sin x{\cos }^{2}x-12{\sin }^2+3{\sin }^{3}x-6\sin x{\cos }^{2}x\\ -6\sin x{\cos }^{2}x+12{\sin }^{3}x\\ {\Delta }^{11}=18\sin x{\cos }^{2}x\\ so,{lim}_{h\to 0}=\frac{18\sin x{\cos }^{2}x}{2}=9\sin x{\cos }^{2}x\end{array}$