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If $A,B,C$ are the angle of a triangle, then show that $\sin A+\sin B-\cos C\le \dfrac32$
I tried substituting $C=180^\circ-(A+B)$ and got stuck. I also tried using the formula $\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$ without any progress. Please help!

user84413
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Hawk
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  • You should use your first hint and note that $\cos(180 - (A+B)) = -\cos(A+B) = \cos A \cos B - \sin A \sin B$. Is that what you did? – user88595 Jan 21 '14 at 19:33
  • yes...that is what I did. – Hawk Jan 21 '14 at 19:37
  • Related : http://math.stackexchange.com/questions/639890/minimum-value-of-cosa-b-cosb-c-cosc-a-is-3-2 and http://math.stackexchange.com/questions/640217/how-to-solve-this-inequality – lab bhattacharjee Jan 22 '14 at 05:05

1 Answers1

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So, we have $\displaystyle \sin\frac{A+B}2=\sin\left(\frac\pi2-\frac C2\right)=\cos\frac C2$

Again, $\displaystyle\cos C=2\cos^2\frac C2-1$

Let $\displaystyle\sin A+\sin B-\cos C=y$

$\displaystyle\implies2\cos\frac C2\cos\frac{A-B}2-\left(2\cos^2\frac C2-1\right)=y$

$\displaystyle\implies2\cos^2\frac C2-2\cos\frac C2\cos\frac{A-B}2+y-1=0\ \ \ \ (1)$ which is a Quadratic Equation in $\cos\frac C2$ which is real as $C$ is

So, the discriminant must be $\ge0 \displaystyle\implies \left(-2\cos\frac{A-B}2\right)^2\ge 4\cdot2\cdot(y-1)$

$\displaystyle\iff y\le 1+\frac{\cos^2\dfrac{A-B}2}2\le 1+\frac12$ as $\displaystyle\cos^2\dfrac{A-B}2\le1$

The equality occurs if $\displaystyle\cos^2\dfrac{A-B}2=1\iff\cos(A-B)=2\cos^2\dfrac{A-B}2-1=1\implies A-B=2n\pi$ where $n$ is an integer

As $0<A,B<\pi, $ it needs $A-B=0\iff A=B$ and will reduces $(1)$ to $\displaystyle2\cos^2\frac C2-2\cos\frac C2+\frac32-1=0$

$\iff \cos\frac C2=\frac12\implies \frac C2=\frac\pi3 $ as $0<\frac C2<\frac\pi2$

$\displaystyle\implies C=\frac{2\pi}3$ and $\displaystyle A=B=\frac{A+B}2=\frac{\pi-C}2=\cdots$

lab bhattacharjee
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