# Prove the given inequality

$$\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C))$$

$$\implies \frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$

Given if $A>B>C$ and $A+B+C=180^\circ$

Is that implication correct if not then please correct it otherwise try to solve the first inequality.

This is not the original problem, but this problem arose when I was solving another trigonometric equation.

Now if anybody prove or disprove the above inequality then that will also be the solution of my problem.

• Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Jan 19, 2014 at 1:36

Yes it is

1. B is always smaller than $90 ^\circ (A > B$ and $180^\circ/2 = 90^\circ )$

2. B is always greater than $0^\circ (B > 0 )$

3. B-C is always greater than $0^\circ (B > C)$

4. B-C is smaller than $90$ ( 1. and $C > 0$ )

So B-C is always between $0^\circ$ and $90^\circ$ ( $90^\circ > (B-C) > 0^\circ$ )

Therefore $\tan(B-C) > 0$

And you may divide both sides in inequalities by the same positive value.

The inequality is equivalent to: $$\sin^2 A \sin(B-C)\cos(A-C) \geq \sin^2 B \sin(A-C)\cos(B-C),$$ or (using twice Briggs' formulas and $\cos(A+B)=-\cos(C)$): $$\sin(3B-A)+\sin(2B+C)+\sin(2A-C)+\sin(2A-3C)\leq \sin(3A-B)+\sin(2A+C)+\sin(2B-C)+\sin(2B-3C)\tag{1},$$ that is equivalent to: $$\sin(2A-2B)\cos(C)\geq\sin(A-B)\cos(4C)\tag{2}$$ or (using the sine duplication formula): $$2\cos(A-B)\cos C\geq \cos(4C).\tag{3}$$ Since $A>B>C>0$ and $A+B+C=\pi$ imply $A-B\leq\pi-3C$, we have: $$2\cos(A-B)\cos C\geq -2\cos C\cos(3C),$$ and equality holds when $A=\pi-2C,B=C$. In this case, $(3)$ holds only if $$2\cos C\cos(3C)+\cos(4C)\leq 0,$$ i.e. only if $C$ is big enough, for istance $$C\geq\operatorname{arccos}\left(\frac{1}{4}\sqrt{7+\sqrt{33}}\right)=0.4679647\ldots>\frac{7\pi}{47}.$$

The original inequality is $\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C))$

And we have to prove this $\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$. Now we have to interchange only the positions of $tan(B-C)$ and $sin^{2}B$. $sin^{2}B$ is always positive since it is a square. We need to care only about $tan(B-C)$.

If $tan(B-C)<0$ then $(B-C)>90^\circ \implies B>90^\circ+C$.

$C$ is a non-zero angle. Therefore $B>90^\circ$ and $A>B>90^\circ$.

Consider the fact that $A+B+C=180^\circ$. This means that the angles A, B , C form the angles of a triangle. But we know that there cannot exist two obtuse angles in a triangle. Therefore our supposition is wrong and hence $B-C<90^\circ$ which leads to $tan(B-C)>0$.

Now we can interchange $tan(B-C)$ and $sin^{2}B$ without changing the inequality sign. Hence we have the given inequality $$\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$

It's correct if $\tan(B-C)>0$, since then you're multiplying both sides by a positive number (the reciprocal of that tangent), unless $\sin^2 B=0$, in which case you've got other problems.

• Can we assume $C > 0$ ? Jan 22, 2014 at 9:43
• @Willemien yes C>0 and $\pi>A>B>C>0$ Jan 22, 2014 at 12:19