In triangle ABC, $\angle C$ is right angle. Find the maximum value of $\cos A.\cos B$.

Question

In triangle ABC, $\angle C$ is right angle. Find the maximum value of $\cos A.\cos B$.

My attempt: $\angle C=\dfrac {\pi}{2}$ then $\angle (A+B)=\dfrac {\pi}{2}$

Let $$y=\cos A.\cos B$$ $$=\cos A.\cos (\dfrac {\pi}{2}-A)$$ $$=\cos A.\sin A$$ So, $y=\dfrac {\sin (2A)}{2}$

Answer 1

Exploring the formulas $$ \cos(x\pm y) = \cos(x)\cos(y) \mp\sin(x)\sin(y), $$we derive $$ \cos(A)\cos(B)= \frac{\cos(A+B)+\cos(A-B)}{2}, $$ and since $\cos(A+B)=\cos(\pi-C)=-\cos(C)$, $$ \cos(A)\cos(B)= \frac{-\cos(C)+\cos(A-B)}{2}, $$ so the maximum is taken when $A-B=0$, that is $A=B=\dfrac{\pi-C}{2}$, and has the value $\dfrac{1-\cos(C)}{2}$.