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Question Number 174837 by mnjuly1970 last updated on 12/Aug/22
sin(A)+ sin(B )+ sin(C)≤ ((3(√3))/2) ≻ Solution ≺ ( A+B +C =π ) l.h.s = 2sin(((A+B)/2) )cos(((A−B)/2))+2sin((C/2))cos((C/2)) = 2cos ((C/2)){2cos(A/2) .cos((B/2) )} = 4cos((A/2)).cos((B/2)).cos((C/2)) l.h.s ≤_(post) ^(previoue) 4 (((3(√3))/8) )=((3(√3))/2) note: { (( I : cos(a)+cos(b)=2cos(((a+b)/2))cos(((a−b)/2)))),(( II: sin(a)+sin(b)=2sin(((a+b)/2))cos(((a−b)/2) ))),(( { (( cos((π/2) −α)=sin(α))),(( sin((π/2) −α)= cos(α))) :})) :}
$$ \\ $$$$\:\:\:\:\:{sin}\left({A}\right)+\:{sin}\left({B}\:\right)+\:{sin}\left({C}\right)\leqslant\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\succ\:\:\:{Solution}\:\prec\:\:\:\:\:\:\left(\:{A}+{B}\:+{C}\:=\pi\:\right) \\ $$$$\:\:\:\:\:\:{l}.{h}.{s}\:=\:\mathrm{2}{sin}\left(\frac{{A}+{B}}{\mathrm{2}}\:\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{{C}}{\mathrm{2}}\right){cos}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left\{\mathrm{2}{cos}\frac{{A}}{\mathrm{2}}\:.{cos}\left(\frac{{B}}{\mathrm{2}}\:\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}{cos}\left(\frac{{A}}{\mathrm{2}}\right).{cos}\left(\frac{{B}}{\mathrm{2}}\right).{cos}\left(\frac{{C}}{\mathrm{2}}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:{l}.{h}.{s}\:\underset{{post}} {\overset{{previoue}} {\leqslant}}\:\mathrm{4}\:\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{note}:\:\begin{cases}{\:{I}\::\:\:{cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)}\\{\:{II}:\:{sin}\left({a}\right)+{sin}\left({b}\right)=\mathrm{2}{sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\:\right)}\\{\:\:\:\begin{cases}{\:{cos}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)={sin}\left(\alpha\right)}\\{\:\:{sin}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)=\:{cos}\left(\alpha\right)}\end{cases}}\end{cases} \\ $$

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