# Inequality involving sine and cosine

I am trying to prove that given $$A,B,C,D,E,F \in \big]0,\frac{\pi}{2}\big]$$ fixed and $$A+C \geq E$$ and the following equation holds for $$\mu = 1$$:

$$\sin(\mu A)\cos(\mu B) + \sin(\mu C)\cos(\mu D) - \sin(\mu E)\cos(\mu F) \geq 0$$ then it holds for all $$\mu\in [0,1]$$.

Note obviously if $$B \leq F$$ and $$D \leq F$$ the statement is trivial to prove.

I checked numerically and it holds, however whatever I tried to prove it did not work.

Any ideas?

• But it is not well defined on the described domain: imagine that $E$ is almost 0. – Fedor Petrov Aug 28 '19 at 15:54
• @FedorPetrov Ah, that is true. I forgot to add another condition. – Loreno Heer Aug 28 '19 at 15:55
• @FedorPetrov I updated the question. – Loreno Heer Aug 28 '19 at 15:58
• I also tried to simplify as: $$\sin(\mu A)\cos(\mu B) + \sin(\mu C)\cos(\mu D) - \sin(\mu \min(A+C, \frac{\pi}{2}))\cos(\mu \arccos(\frac{\sin(A)\cos(B)+\sin(C)\cos(D)}{\sin(\min(A+C, \frac{\pi}{2}))}))$$, but showing that this is smaller than the above and also showing it is still larger than zero does not look like it is easier. – Loreno Heer Aug 30 '19 at 9:53
• may I ask, does this question come from certain geometry problem? – Fedor Petrov Sep 3 '19 at 9:09

I am afraid it is false. Take $$F=0$$, $$A=C$$, $$E=2A$$, $$\mu=1/2$$. Then we are given $$\cos B+\cos D=2\cos A$$ and should prove $$\cos B/2+\cos D/2\geqslant 2\cos A/2$$. But if $$\cos B=x$$, then $$\cos B/2=\sqrt{(1+x)/2}$$, this function is concave, thus inverse inequality $$\cos B/2+\cos D/2\leqslant 2\cos A/2$$ holds.
• Please note that I specified $A,B,\ldots,F \in ]0,\frac{\pi}{2}]$. In particular $F=0$ is not possible. – Loreno Heer Sep 2 '19 at 10:40
• Yes, but it is certainly not important: take very small $F$ if you do not like zero, and accordingly slightly change $B,D$. – Fedor Petrov Sep 2 '19 at 11:13
• I am not sure I fully understand your example. I see how you derive the first equality, however I believe it should be $\geq$ there. The last inequality does not contradict the required one as there could be equality... – Loreno Heer Sep 2 '19 at 11:31
• It is strict if $B\ne D$. – Fedor Petrov Sep 2 '19 at 12:09