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If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$

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What did you try ? –  Claude Leibovici Aug 24 at 13:31
    
I think the maximum is at $A=B=C=\frac{5}{12}\pi$, because $\sin x$ is concave on $[0,\frac 12 \pi]$. –  Ragnar Aug 24 at 13:33
    
I think it occurs for $A=B=C=\frac{5\pi}{12}$ but can't justify. –  Pankaj Sinha Aug 24 at 13:34
    
@Ragnar.But there is no such condition. In fact $x\in(0,\frac{5\pi}{4})$. –  Pankaj Sinha Aug 24 at 13:38

5 Answers 5

The function $x\mapsto\sin x$ is concave on the interval $[0,\pi]$. It follows that when all three of $A$, $B$, $C$ are in this interval we have $${1\over3}(\sin A+\sin B+\sin C)\leq\sin{A+B+C\over3}=\sin{5\pi\over12}\ .$$ It follows that in this case $$\sin A+\sin B+\sin C\leq 3\sin{5\pi\over12}={3(1+\sqrt{3})\over 2\sqrt{2}}\doteq2.8978\ .\tag{1}$$ When, e.g., $\pi<A<{5\pi\over4}$ then $B+C<{\pi\over4}$, and one gets $$\sin A+\sin B+\sin C<\sin B+\sin C\leq 2\sin{B+C\over2}<2\sin{\pi\over8}\doteq0.765\ .$$ The conclusion is that the value exhibited in $(1)$ is indeed maximal.

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Maybe the inequality $$|\sin x| \leq |x|$$ helps.

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$\sin{A}+\sin{B}+\sin{C}=2\sin{\frac{A+B}{2}}\cos{\frac{A-B}{2}}+\sin{C}\le 2\sin{\frac{A+B}{2}}+\sin{(\frac{5\pi}{4}-(A+B))}.$ Equality holds when $A=B$. Thus we consider the function $$f(x)=2\sin{x}+\sin{\left(\frac{5\pi}{4}-2x\right)}=2\sin{x}+\sin{\left(2x-\frac{\pi}{4}\right)}.$$ defined on $(0,\frac{5\pi}{8})$. Then $$f'(x)=2\cos{x}+2\cos{\left(2x-\frac{\pi}{4}\right)}$$ It achieves zero when $x=2x+\frac{3\pi}{4}+2k\pi$ or $x=2k\pi-2x+\frac{5\pi}{4}$. That is, $0<x<\frac{5\pi}{12},f'>0$ and $5\pi/12<x<5\pi/8,f'<0$. Thus $f$ achieves maximum at $x=5\pi/12$. Namely, $A=B=C=5\pi/12$.

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From a purely algebraic point of view, elimating $C$ from the constraint leads to maximizing $$F=\sin \left(A+B-\frac{\pi }{4}\right)+\sin (A)+\sin (B)$$ Taking derivatives $$F'_A=\cos \left(A+B-\frac{\pi }{4}\right)+\cos (A)=0$$ $$F'_B=\cos \left(A+B-\frac{\pi }{4}\right)+\cos (B)=0$$ This implies $A=B$ and then $$\cos(2A-\frac{\pi }{4})+\cos(A)=2 \cos \left(\frac{A}{2} -\frac{\pi }{8}\right) \cos \left(\frac{3A}{2} -\frac{\pi }{8}\right)=0$$ and the only acceptable solution is $A=\frac{5 \pi }{12}$ with $B=C=A$ and then a maximum value equal to $$\frac{3 \left(1+\sqrt{3}\right)}{2 \sqrt{2}}$$

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Also check the boundary condition $A=0$, but then the sum is less than 2. –  Michael Aug 24 at 14:34
    
@Michael. You are perfectly correct. Cheers :-) –  Claude Leibovici Aug 24 at 14:36

We can use the method of Lagrange Multipliers. Note that the constraint function is $$ g(A,B,C) = A+B+C $$ and the maximization function is $$ f(A,B,C) = \sin A + \sin B + \sin C. $$We can find $ \nabla f $ and $ \nabla g $ and we get $$ \begin {eqnarray*} \nabla g &=& \left< 1, 1, 1 \right>, \\ \nabla f &=& \left< \cos A, \cos B, \cos C \right>. \end {eqnarray*} $$ Then, setting $ \nabla f \propto \nabla g $, we get $ \cos A = \cos B = \cos C $. We can go back to the restrictions that $A,B,C>0$ and $A+B+C=\frac{5\pi}{4}$ and we see that the only solution here is $$ (A,B,C) = \left( \frac{5\pi}{12}, \frac{5\pi}{12}, \frac{5\pi}{12} \right), $$ at which $ f(A,B,C) = \boxed{\frac {3 \sqrt{2+\sqrt{3}}}{2}} $. $\blacksquare$

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