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Math Help - a trigonometric inequality

  1. #1
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    a trigonometric inequality

    ABC-acute triangle. prove that sinA+sinB>cosA+cosB+cosC
    thanks
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  2. #2
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    Quote Originally Posted by ely_en View Post
    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
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    \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
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  4. #4
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    Hello,
    Quote Originally Posted by ely_en View Post
    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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