In $\triangle ABC$, if $\angle B=60^\circ$ and $\sin A\sin C=\lambda$, then what is the range of $\lambda$? In $\triangle ABC$ if $\angle B=60^\circ$ and $\sin A\sin C=\lambda$, then what is the range of $\lambda$?


Try: Using $$\cos B=\frac{a^2+c^2-b^2}{2ab}$$

So $a^2-ac+c^2=b^2$. From $\sin A\sin C=\lambda$, substitute $$ac=4R^2\lambda$$
where $R$ is circumradius of $\triangle ABC$.

Could some help me how to calculate $\lambda$? Thanks.
$$\sin A\sin C=\frac12\bigg(\cos(A-C)-\cos(A+C))=\frac12\bigg(\cos D+\frac12\bigg)$$ where $D=|A-C|$ which lie between $0^\circ$ to $120^\circ$
So $-\frac{\sqrt3}2 \leq \cos D \leq 1$

So the range is from $\bigg[\frac{1-\sqrt3}4, \frac34\bigg]$
$$\sin A\sin C=\frac12\bigg(\cos(A-C)-\cos(A+C))=\frac12\bigg(\cos D+\frac12\bigg)$$ where $D=|A-C|$ which lie between $0^\circ$ to $120^\circ$
So $-\frac{\sqrt3}2 \leq \cos D \leq 1$

So the range is from $\bigg[\frac{1-\sqrt3}4, \frac34\bigg]$