a trigonometric inequality

**ely_en** · September 1st 2008, 04:52 AM

ABC-acute triangle. prove that sinA+sinB>cosA+cosB+cosC
thanks

**NonCommAlg** · September 1st 2008, 09:20 PM

Originally Posted by ely_en

ABC-acute triangle. prove that sinA+sinB>cosA+cosB+cosC
thanks

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

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

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

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

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

**ely_en** · September 2nd 2008, 12:06 AM

Originally Posted by NonCommAlg

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

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

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

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

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

could you tell me how did you get $\frac{\sqrt{2}\cos \left(\frac{A-B}{2} \right)}{\cos \left(\frac{\pi}{4} - \frac{C}{2} \right)}\cos C$ from $2\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{C}{2} \right) \cos \left(\frac{A-B}{2} \right)$ , please?

Thank you very much

**Moo** · September 2nd 2008, 12:16 AM

Hello,

Originally Posted by ely_en

could you tell me how did you get $\frac{\sqrt{2}\cos \left(\frac{A-B}{2} \right)}{\cos \left(\frac{\pi}{4} - \frac{C}{2} \right)}\cos C$ from $2\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{C}{2} \right) \cos \left(\frac{A-B}{2} \right)$ , please?

Thank you very much

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

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

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

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

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