If $\cos B\cos C +\sin B\sin C \sin^2A=1$, then triangle ABC is what kind of triangle?

Question

$$\cos B \cos C +\sin B \sin C - \sin B \sin C \cos^2A=1$$

$$\cos (B-C)-\sin B \sin C \cos^2(B+C)=1$$

$$\cos (B-C)-\frac 12 (\cos (B-C)-\cos (B+C))\cos^2(B+C)=1$$

How do I proceed from here?

Answer 1

$$\cos(B-C)-\frac 12\big(\cos(B-C)-\cos(B+C)\big)\cos^2A=1 \\ \implies \cos(B-C) -\frac 12\big(\cos(B-C)+\cos A\big) \cos^2A =1 \\ \implies2\cos(B-C)-\cos(B-C)\cos^2A-\cos^3A=2 \\ \implies \cos(B-C)[1+\sin^2A]=2+\cos^3A \\\implies\cos(B-C)=\frac{2+\cos^3A}{1+\sin^2A} $$ The RHS must be $\le 1$, i.e $\cos^3A-\cos^2A+1\ge 1$.

Consider a function $f(x)=x^3-x^2+1$ on $[-1,1]$. We have $f(-1)=-1, f(0)=1, f(1)=1$ and $f’(x)=3x^2-2x=0\implies x=0,\frac 23$ corresponding to a maximum and minimum respectively. This means $f(x)\le 1$ with equality only at $x=0,1$.

Therefore, to make our inequality hold we must have either $\cos A=0$ or $\cos A=1$, but $A\ne 0$ so $\cos A=0 \implies \boxed{A=\frac{\pi}{2}}$

So, $\cos(B-C)=1 \implies \boxed{B=C=\frac{\pi-\frac{\pi}{2}}{2}=\frac{\pi}{4}}$

Answer 2

As $0<\sin A,\sin B,\sin C<1,$

For real $n\ge1,$

$$\cos B\cos C+\sin B\sin C\sin^nA\le\cos B\cos C+\sin B\sin C$$

$$\implies1\le\cos(B-C)$$ which is possible only if

$$\cos(B-C)=1\text{ and }(\sin^n A=1\iff\sin A=1\text{ as }\sin A>0)$$