# a trigonometric inequality

• Sep 1st 2008, 04:52 AM
ely_en
a trigonometric inequality
ABC-acute triangle. prove that sinA+sinB>cosA+cosB+cosC
thanks
• Sep 1st 2008, 09:20 PM
NonCommAlg
Quote:

Originally Posted by ely_en
ABC-acute triangle. prove that sinA+sinB>cosA+cosB+cosC
thanks

$\displaystyle \boxed{1}: \ \ \sqrt{2}\cos \left(\frac{A-B}{2}\right) > \cos \left(\frac{\pi}{4}-\frac{C}{2}\right).$

Proof: since A, B and C are between 0 and $\displaystyle \frac{\pi}{2},$ both sides of the inequality are positive. now: $\displaystyle 2\cos^2 \left(\frac{A-B}{2}\right)=1+\cos(A-B) > 1 > \cos^2 \left(\frac{\pi}{4}-\frac{C}{2}\right). \ \ \ \square$

$\displaystyle \boxed{2} \ \ \sin A +\sin B > \cos A + \cos B + \cos C.$

Proof: $\displaystyle \sin A - \cos A + \sin B - \cos B= \sqrt{2}\left(\sin \left(A-\frac{\pi}{4} \right) + \sin \left(B - \frac{\pi}{4} \right) \right)$

$\displaystyle =2\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{C}{2} \right) \cos \left(\frac{A-B}{2} \right)=\frac{\sqrt{2}\cos \left(\frac{A-B}{2} \right)}{\cos \left(\frac{\pi}{4} - \frac{C}{2} \right)}\cos C > \cos C,$ by $\displaystyle \boxed{1}. \ \ \ \ \ \ \square$
• Sep 2nd 2008, 12:06 AM
ely_en
Quote:

Originally Posted by NonCommAlg
$\displaystyle \boxed{1}: \ \ \sqrt{2}\cos \left(\frac{A-B}{2}\right) > \cos \left(\frac{\pi}{4}-\frac{C}{2}\right).$

Proof: since A, B and C are between 0 and $\displaystyle \frac{\pi}{2},$ both sides of the inequality are positive. now: $\displaystyle 2\cos^2 \left(\frac{A-B}{2}\right)=1+\cos(A-B) > 1 > \cos^2 \left(\frac{\pi}{4}-\frac{C}{2}\right). \ \ \ \square$

$\displaystyle \boxed{2} \ \ \sin A +\sin B > \cos A + \cos B + \cos C.$

Proof: $\displaystyle \sin A - \cos A + \sin B - \cos B= \sqrt{2}\left(\sin \left(A-\frac{\pi}{4} \right) + \sin \left(B - \frac{\pi}{4} \right) \right)$

$\displaystyle =2\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{C}{2} \right) \cos \left(\frac{A-B}{2} \right)=\frac{\sqrt{2}\cos \left(\frac{A-B}{2} \right)}{\cos \left(\frac{\pi}{4} - \frac{C}{2} \right)}\cos C > \cos C,$ by $\displaystyle \boxed{1}. \ \ \ \ \ \ \square$

could you tell me how did you get $\displaystyle \frac{\sqrt{2}\cos \left(\frac{A-B}{2} \right)}{\cos \left(\frac{\pi}{4} - \frac{C}{2} \right)}\cos C$ from $\displaystyle 2\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{C}{2} \right) \cos \left(\frac{A-B}{2} \right)$, please?(Thinking) Thank you very much
• Sep 2nd 2008, 12:16 AM
Moo
Hello,
Quote:

Originally Posted by ely_en
could you tell me how did you get $\displaystyle \frac{\sqrt{2}\cos \left(\frac{A-B}{2} \right)}{\cos \left(\frac{\pi}{4} - \frac{C}{2} \right)}\cos C$ from $\displaystyle 2\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{C}{2} \right) \cos \left(\frac{A-B}{2} \right)$, please?(Thinking) Thank you very much

Are you familiar with $\displaystyle \sin(2x)=2 \sin(x) \cos(x)$ ?

$\displaystyle 2 \sin \left(\tfrac \pi 4-\tfrac C2\right) \cos \left(\tfrac \pi 4-\tfrac C2\right)=\sin \left(\tfrac \pi 2-C \right)$

But we know that $\displaystyle \sin \left(\tfrac \pi 2-x\right)=\cos(x)$

Therefore $\displaystyle 2 \sin \left(\tfrac \pi 4-\tfrac C2\right) \cos \left(\tfrac \pi 4-\tfrac C2\right)=\cos(C) \quad \blacksquare$