Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$

Question

To prove $$\sin{A}+\sin{B}-\cos{C}\le\frac32$$
Given $A+B+C=\pi$ and $A,B,C>0$

I have managed to convert LHS to $$1-4\cos{\frac C2}\sin{\frac{A+C-B}2}\cos{\frac{A-B-C}2}$$ but that clearly isn't very helpful

One other conversion was $$\sin A+\sin B+\cos A\cos B-\sin A\sin B$$ but not sure how to proceed.

Any hints will be appreciated

Duplicate found instantly with Approach0 – Martin R Feb 13 '18 at 08:41 — Martin R, Feb 13 '18 at 08:41

score 3 · Accepted Answer · answered Feb 13 '18 at 07:00

3

Let $y=\sin A+\sin B-\cos C=2\cos\dfrac C2\cos\dfrac{A-B}2-2\cos^2\dfrac C2+1$

$$\iff2\cos^2\dfrac C2-2\cos\dfrac C2\cos\dfrac{A-B}2+y-1=0$$

As $\cos\dfrac C2$ is real,the discriminant $\ge0$

$$4\cos^2\dfrac{A-B}2-8(y-1)\ge0\iff8y\le8+4\cos^2\dfrac{A-B}2\le8+4$$

answered Feb 13 '18 at 07:00

lab bhattacharjee

So, the equality occurs if $C=\dfrac{2\pi}3, A=B=?$ – lab bhattacharjee Feb 13 '18 at 07:07
$\pi/6$, how did you get $2\pi/3$ though? – Anvit Feb 13 '18 at 07:13
@AFalseName, For the equality, $$\cos^2\dfrac{A-B}2=?$$ – lab bhattacharjee Feb 13 '18 at 07:14
=1 implying A=B, But how C = $2\pi/3$ – Anvit Feb 13 '18 at 07:15
oh, did you solve the first eq by plugging C = ($\pi-2A$) – Anvit Feb 13 '18 at 07:18
@AFalseName, If $A=B,y=\dfrac32$ $$0=2\cos^2\dfrac C2-2\cos\dfrac C2\cos\dfrac{A-B}2+y-1=?$$ – lab bhattacharjee Feb 13 '18 at 07:19
Yep, Got it. thank you – Anvit Feb 13 '18 at 07:20
Note that you answered an identical question before: https://math.stackexchange.com/a/647141. – Martin R Feb 13 '18 at 08:36
@MartinR, Thanks. I forgot the exact match. I've already linked with some related ones in the post itself. – lab bhattacharjee Feb 13 '18 at 08:46

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