# Geometric meaning of trigonometric relations

According to a paper by Zhiqin Lu in the Mathematical Gazette (the British publication, not the Boston-area newsletter, if that still exists (or even if it doesn't)) in 2007(?), if $u+v+w=\pi$ and $a,b,c \ge 0$, then $a + b + c \ge 2 \sqrt{bc} \cos u + 2 \sqrt{ca} \cos v + 2 \sqrt{ab} \cos w$. It wasn't hard to show that equality holds iff $a:b:c = \sin^2 u : \sin^2 v : \sin^2 w$, but if I recall correctly, that wasn't in the paper. I would think that there must be some natural geometric interpretation of this proposition. What is it?

The case where equality holds says that if $u+v+w = \pi$, then $\sin^2 u + \sin^2 v + \sin^2 w = 2 \sin u \sin v \cos w + 2 \sin u \cos v \sin w + 2 \cos u \sin v \sin w$. That one has a simple geometric interpretation as a sort of mash-up of the law of sines and the law of cosines. But now suppose we say that if $$\sum_i u_i =\pi$$ then $$\sum_{i=1}^\infty \sin^2 u_i = \sum_{\text{even }n\ge 2} (-1)^{(n-2)/2}n\sum_{|A|=n} \prod_{i\in A}\sin u_i \prod_{i\not\in A} \cos u_i$$ The case of a sum of three variables that add up to $\pi$ is the special case in which all but three of these are 0. Does it have a geometric meaning?

Later edit: So far we have an answer about the inequality, but not yet about the equality. I will probably comment soon on the former.

• Michael: the Boston-area Mathematical Gazette still exists. I got an email about the latest issue last week. You can check it out online at wpi.edu/academics/Depts/Math/News/gazette.html. Apr 25 '11 at 5:16
• I have posted a proof of the inequality on MathLinks some years ago, and IIRC it reduced it to a $\left(\text{some distance}\right)^2\geq 0$. Can anyone find it? Apr 25 '11 at 10:00
• Found it. See my answer below. Apr 25 '11 at 10:21
• To say that the square of some particular distance is nonnegative tells us only what we would know without knowing which particular distance it is. Apr 25 '11 at 19:12

In an ancient MathLinks topic (post #6; but see below for a copy) I have given a proof of the inequality by reducing it to $$\left(\sqrt{a}\vec{p}+\sqrt{b}\vec{q}+\sqrt{c}\vec{r}\right)^2\geq 0$$, where multiplication means scalar product of vectors and $$\vec{p}$$, $$\vec{q}$$, $$\vec{r}$$ are unit length vectors chosen in such a way that the angles between them are $$\pi-u$$, $$\pi-v$$, $$\pi-w$$, respectively. This rewrites geometrically as follows: Pick a point $$P$$ in the plane, and take three points $$A$$, $$B$$, $$C$$ such that $$PA=\sqrt{a}$$, $$PB=\sqrt{b}$$, $$PC=\sqrt{c}$$, $$\measuredangle BPC=\pi-u$$, $$\measuredangle CPA=\pi-v$$ and $$\measuredangle APB=\pi-w$$. Then, the difference between the left hand side and the right hand side of your inequality is $$9$$ times the square of the distance between the point $$P$$ and the centroid of triangle $$ABC$$. Equality thus holds if and only if $$P$$ is the centroid of triangle $$ABC$$; this is equivalent to the assertion that the triangles $$BPC$$, $$CPA$$, $$APB$$ have equal areas; this, in turn, is equivalent to the assertion that $$\sqrt{a}:\sqrt{b}:\sqrt{c}=\sin u:\sin v:\sin w$$ (because the area of triangle $$BPC$$ is $$\frac{1}{2}\cdot PB\cdot PC\cdot \sin\measuredangle BPC=\frac{1}{2}\sqrt{b}\sqrt{c}\sin u$$ etc.).

For better searchability, let me copy my MathLinks posts over here (finding some old post on MathLinks is almost impossible as for now). Note that I do not claim originality for the theorems.

Theorem 1. Let $$x$$, $$y$$, $$z$$ be three real numbers and $$A$$, $$B$$, $$C$$ three real angles such that $$A + B + C = 180^{\circ}$$. Then,

$$x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$$.

Proof of Theorem 1. We will denote by $$\measuredangle\left(\overrightarrow{p};\;\overrightarrow{q}\right)$$ the directed angle between two vectors $$\overrightarrow{p}$$ and $$\overrightarrow{q}$$ (note that this is a directed angle modulo $$360^{\circ}$$).

For any two vectors $$\overrightarrow{p}$$ and $$\overrightarrow{q}$$, we are going to denote by $$\overrightarrow{p}\cdot\overrightarrow{q}$$ the scalar product of the vectors $$\overrightarrow{p}$$ and $$\overrightarrow{q}$$.

For any vector $$\overrightarrow{p}$$, we are going to denote by $$\overrightarrow{p}^2$$ the scalar product $$\overrightarrow{p}\cdot\overrightarrow{p}$$. Every vector $$\overrightarrow{p}$$ satisfies $$\overrightarrow{p}^2=\left|\left|\overrightarrow{p}\right|\right|^2\geq 0$$.

Let $$\overrightarrow{a}$$ be a vector of unit length. Let $$\overrightarrow{b}$$ be a vector of unit length such that $$\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)=180^{\circ}-C$$. Let $$\overrightarrow{c}$$ be a vector of unit length such that $$\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$$. Then,

$$\measuredangle\left(\overrightarrow{c};\;\overrightarrow{a}\right)=360^{\circ}-\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)-\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)$$ $$=360^{\circ}-\left(180^{\circ}-C\right)-\left(180^{\circ}-A\right)=C+A=180^{\circ}-B$$

(since $$A + B + C = 180^\circ$$).

Now, all the vectors $$\overrightarrow{a}$$, $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$ have unit length: $$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1$$. Thus, $$\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$$ yields

$$\overrightarrow{b}\cdot\overrightarrow{c}=\left|\overrightarrow{b}\right|\cdot\left|\overrightarrow{c}\right|\cdot\cos\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=1\cdot 1\cdot\cos\left(180^{\circ}-A\right)=\cos\left(180^{\circ}-A\right)$$ $$=-\cos A$$.

Similarly, we obtain $$\overrightarrow{c}\cdot\overrightarrow{a}=-\cos B$$ and $$\overrightarrow{a}\cdot\overrightarrow{b}=-\cos C$$. Thus,

$$\left(x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}+z\cdot\overrightarrow{c}\right)^2$$

$$=\left(x\cdot\overrightarrow{a}\right)^2+\left(y\cdot\overrightarrow{b}\right)^2+\left(z\cdot\overrightarrow{c}\right)^2$$

$${}+2\cdot y\cdot\overrightarrow{b}\cdot z\cdot\overrightarrow{c}+2\cdot z\cdot\overrightarrow{c}\cdot x\cdot\overrightarrow{a}+2\cdot x\cdot\overrightarrow{a}\cdot y\cdot\overrightarrow{b}$$

$$=x^2\underbrace{\cdot\left|\overrightarrow{a}\right|^2}_{=1^2}+y^2\cdot\underbrace{\left|\overrightarrow{b}\right|^2}_{=1^2}+z^2\cdot\underbrace{\left|\overrightarrow{c}\right|^2}_{=1^2}+2yz\cdot\underbrace{\overrightarrow{b}\cdot\overrightarrow{c}}_{=-\cos A}+2zx\cdot\underbrace{\overrightarrow{c}\cdot\overrightarrow{a}}_{=-\cos B}+2xy\cdot\underbrace{\overrightarrow{a}\cdot\overrightarrow{b}}_{=-\cos C}$$

$$=x^2\cdot 1^2+y^2\cdot 1^2+z^2\cdot 1^2+2yz\cdot\left(-\cos A\right)+2zx\cdot\left(-\cos B\right)+2xy\cdot\left(-\cos C\right)$$

$$=x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C$$.

Since we, obviously, have $$\left(x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}+z\cdot\overrightarrow{c}\right)^2\geq 0$$, we thus get $$x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C\geq 0$$, so that $$x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$$, and Theorem 1 is proven.

Other proofs of Theorem 1 can be found at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5243 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42509 .

Theorem 1 is equivalent to the following, also quite useful (for olympiad mathematics and magazine problem sections, that is, although I would not be surprised to see more applications) inequality:

Theorem 2. Let $$x$$, $$y$$, $$z$$ be three real numbers and $$A$$, $$B$$, $$C$$ three real angles such that $$A + B + C$$ is a multiple of $$180^{\circ}$$. Then,

$$\left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$$.

We will only show a proof of Theorem 2 using Theorem 1: First, we can WLOG assume that $$A + B + C = 180^{\circ}$$. This is because the inequality

$$\left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$$

will not change if we add a multiple of 180° to one of the angles A, B and C (because $$\sin^{2}\left(180^{\circ} + u\right) = \sin^{2}u$$ for every u), and consequently, since A + B + C is a multiple of 180°, we can add a multiple of 180° to the angle A such that, after this, we will have A + B + C = 180°.

Now, for A + B + C = 180°, we have

$$\left(180^{\circ} - 2A\right) + \left(180^{\circ} - 2B\right) + \left(180^{\circ} - 2C\right) = 540^{\circ} - 2\cdot\left(A + B + C\right)$$ $$= 540^{\circ} - 2\cdot 180^{\circ} = 180^{\circ}$$.

Hence, Theorem 1 (applied to $$180^{\circ}-2A$$, $$180^{\circ}-2B$$, $$180^{\circ}-2C$$ instead of $$A$$, $$B$$, $$C$$) yields

$$x^{2} + y^{2} + z^{2}\geq 2yz\cos\left(180^{\circ} - 2A\right) + 2zx\cos\left(180^{\circ} - 2B\right) + 2xy\cos\left(180^{\circ} - 2C\right)$$.

Since $$\cos\left(180^{\circ} - 2A\right) = - \cos\left(2A\right) = - \left(1 - 2\sin^{2}A\right) = 2\sin^{2}A - 1$$ and similarly $$\cos\left(180^{\circ} - 2B\right) = 2\sin^{2}B - 1$$ and $$\cos\left(180^{\circ} - 2C\right) = 2\sin^{2}C - 1$$, this becomes

$$x^{2} + y^{2} + z^{2}\geq 2yz\left(2\sin^{2}A - 1\right) + 2zx\left(2\sin^{2}B - 1\right) + 2xy\left(2\sin^{2}C - 1\right)$$ $$\Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right) - \left(2yz + 2zx + 2xy\right)$$ $$\Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2} + \left(2yz + 2zx + 2xy\right)\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$$ $$\Longleftrightarrow\ \ \ \ \ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$$,

and Theorem 2 is proven.

Theorem 2 also trivially follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 and was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3849 ...