# Find the maximum value of $T=\frac{2}{3}(\cos 2A-\cos 2B)-\tan\frac{C}{2}$

Let $ABC$ be a triangle. Find the maximum value of $$T=\frac{2}{3}(\cos 2A-\cos 2B)-\tan\frac{C}{2}$$

Please give me some hints. I don't know where to start Thanks

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Are $A,B,C$ angles of a triangle? –  user17762 Apr 24 '13 at 18:57
yes, Let ABC be a triangle –  septimus Apr 24 '13 at 18:58
sorry is cos2A-cos2B –  septimus Apr 24 '13 at 19:06
Are you allowed to use derivatives? –  Américo Tavares Apr 24 '13 at 19:53
yes i am allowed to use derivatives. –  septimus Apr 24 '13 at 19:55

Hint.

$$T=-\frac43\sin(A+B)\sin(A-B)-\frac{1}{\tan\frac{A+B}{2}}.$$

If $A+B\ge \pi/2$, $$T\le \frac43\sin(A+B)-\frac{1}{\tan\frac{A+B}{2}},$$

where equality holds for $A-B=-\pi/2$.

If $A+B<\pi/2$,

$$T\le \frac43\sin^2(A+B)-\frac{1}{\tan\frac{A+B}{2}},$$

where equality holds for $A=0$.

Note that if you set $t=\tan\frac{A+B}{2}$, $$\sin (A+B)=\frac{2t}{1+t^2}.$$ Use derivatives with respect to $t$.

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