Finding Maximum

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Finding Maximum

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Source: From my friend

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FUSK

#1 h Mar 25, 2020, 9:05 AM

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Let $ABC$ be a triangle.Find the maximum of $sinA+sinBsin{\frac{C}{2}}$

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#2 h Mar 25, 2020, 9:31 AM • 1 Y

Y by Mango247

I have found this here
$\cos \left(\frac{1}{2} \left(4 \tan ^{-1}\left(\frac{1}{5 \sqrt{\frac{2}{174+16 \sqrt{21}-25 \sqrt{\frac{33152}{625}+\frac{5568 \sqrt{21}}{625}}}}}\right)+4 \tan ^{-1}\left(\frac{583925760 \sqrt{87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}}-\frac{1560674304}{5} \left(87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}\right)^{3/2}+\frac{2473721856}{125} \left(87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}\right)^{5/2}-\frac{605159424 \left(87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}\right)^{7/2}}{3125}}{-\frac{4426478993276 928}{3125}-\frac{1042436327473152 \sqrt{21}}{3125}+\frac{48781030588416 \sqrt{518+87 \sqrt{21}}}{3125}+\frac{168510357504 \left(518+87 \sqrt{21}\right)^{3/2}}{3125}+\frac{21966140473344 \sqrt{21 \left(518+87 \sqrt{21}\right)}}{3125}}\right)\right)\right)$

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FUSK

#3 h Mar 25, 2020, 9:43 AM

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Dr Sonnhard Graubner wrote:

I have found this here
$\cos \left(\frac{1}{2} \left(4 \tan ^{-1}\left(\frac{1}{5 \sqrt{\frac{2}{174+16 \sqrt{21}-25 \sqrt{\frac{33152}{625}+\frac{5568 \sqrt{21}}{625}}}}}\right)+4 \tan ^{-1}\left(\frac{583925760 \sqrt{87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}}-\frac{1560674304}{5} \left(87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}\right)^{3/2}+\frac{2473721856}{125} \left(87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}\right)^{5/2}-\frac{605159424 \left(87+8 \sqrt{21}-4 \sqrt{518+87 \sqrt{21}}\right)^{7/2}}{3125}}{-\frac{4426478993276 928}{3125}-\frac{1042436327473152 \sqrt{21}}{3125}+\frac{48781030588416 \sqrt{518+87 \sqrt{21}}}{3125}+\frac{168510357504 \left(518+87 \sqrt{21}\right)^{3/2}}{3125}+\frac{21966140473344 \sqrt{21 \left(518+87 \sqrt{21}\right)}}{3125}}\right)\right)\right)$

Please show me your solution.(And don't use calculator.)
And the answer is $\frac{\sqrt{7}}{2}$ .I don't know if your answer is equal to this result.

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sqing

#5 h Mar 25, 2020, 11:30 AM

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Good.
Let $ABC$ be a triangle.Prove that $sinA+sinBsin{\frac{C}{2}}\leq \frac{\sqrt{7}}{2}$

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sqing

#6 h Mar 25, 2020, 12:05 PM

Y by

The following inequality is easy.
Let $ABC$ be a triangle.Prove that $sinA+sinBsinC\leq \frac{\sqrt{5}+1}{2}$ $cosA+sinBcos C< \frac{\sqrt{5}+1}{2}$ $sinA+sinBcosC<2$ $cosA+cosBcosC\leq 1$ $cos{A}+ 2cos{B} cos{C} \le 1$ $sinA- 2sinBcosC\le 1$ h
Let $ABC$ be a triangle.Find the maximum of $sinA+sinBcosC$ , $cosA+sinBcosC$ , $cosA+cosBcos{\frac{C}{2}}$ , $sinA+sinBcos{\frac{C}{2}}$ and $cosA+sinBcos{\frac{C}{2}}.$

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#8 h Mar 25, 2020, 1:17 PM

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Let $f(x,y)=\sin(x)+\sin(y)\cos\left(\frac{x+y}{2}\right)$ then you have to solve
$\frac{\partial f(x,y)}{\partial x}=\cos(x)-\frac{1}{2}\sin(y)\sin\left(\frac{x+y}{2}\right)=0$ and
$\frac{\partial f(x,y)}{\partial y}=\cos(y)\cos\left(\frac{x+y}{2}\right)-\frac{1}{2}\sin(y)\sin\left(\frac{x+y}{2}\right)=0$ and $0\le x,y\le \pi$

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FUSK

#9 h Mar 26, 2020, 7:55 AM

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Dr Sonnhard Graubner wrote:

Let $f(x,y)=\sin(x)+\sin(y)\cos\left(\frac{x+y}{2}\right)$ then you have to solve
$\frac{\partial f(x,y)}{\partial x}=\cos(x)-\frac{1}{2}\sin(y)\sin\left(\frac{x+y}{2}\right)=0$ and
$\frac{\partial f(x,y)}{\partial y}=\cos(y)\cos\left(\frac{x+y}{2}\right)-\frac{1}{2}\sin(y)\sin\left(\frac{x+y}{2}\right)=0$ and $0\le x,y\le \pi$

Please provide an elementary solution.
I am just a high school student and not a college student.

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sqing

#10 h Apr 6, 2025, 3:40 AM

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sqing wrote:

Let $ABC$ be a triangle.Prove that $sinA+sinBsin{\frac{C}{2}}\leq \frac{\sqrt{7}}{2}$

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sqing

#11 h Apr 6, 2025, 3:41 AM

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In a triangle ABC. Prove that $sinA+sinBcos\frac{C}{2}<2$

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