Question

Prove tha in a triangle $ABC$, if $\cos A\cos 2B+\sin A\sin 2B\sin C=1$, then
$A,B,C$ are in A.P.
$B,A,C$ are in A.P.
$\frac{r}{R}=2$

Answer 1

$\cos A\cos 2B+\sin A\sin 2B\sin C=1------\left(1\right)$

$\sin C=\frac{1-\cos A\cos 2B}{\sin A\sin 2B}$

$\sin C\le 1\Rightarrow \frac{1-\cos A\cos 2B}{\sin A\sin 2B}\le 1$

$1-\cos A\cos 2B\le \sin A\sin 2B$

$1\le \cos A\cos 2B+\sin A\sin 2B$

$1\le \cos (2B-A)$

This is only possible when

$\cos (2B-A)=1$

Thus $2B-A=0$

$A=2B$

Putting $2B=A$ in eqn $\left(1\right)$

${\cos }^{2}A+{\sin }^{2}A\sin C=1$

${\sin }^{2}A\sin C=1-{\cos }^{2}A$

${\sin }^{2}A(1-\sin C)=0$

Thus $\sin C=1$

$C=\pi /2$

$A+B+C=\pi$

$A+B=\pi /2$

$3B=\pi /2$

$B=\pi /6$

$A=\pi /3$

$C=\pi /2$

Thus $C-A=A-B$

Therefore $B,A,C$ are in $AP$