# cos/sin inequality

Here's a problem from the 2004 Russian Mathematical Olympiad [Lecture Notes, p 61]:

Let \(a,b,c\) be positive numbers, satisfying \(\displaystyle a+b+c=\frac{\pi}{2}\), prove that

\( \mbox{cos}(a)+\mbox{cos}(b)+\mbox{cos}(c)\gt \mbox{sin}(a)+\mbox{sin}(b)+\mbox{sin}(c) \)

### References

- Xu Jiagu,
*Lecture Notes on Mathematical Olympiad Courses*, v 8, (For senior section, v 1), World Scientific, 2012

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Copyright © 1996-2017 Alexander Bogomolny

Let \(a,b,c\) be positive numbers, satisfying \(\displaystyle a+b+c=\frac{\pi}{2}\), prove that

\( \mbox{cos}(a)+\mbox{cos}(b)+\mbox{cos}(c)\gt \mbox{sin}(a)+\mbox{sin}(b)+\mbox{sin}(c) \)

It appears that the problem admits a brute-force, a rather straightforward, solution (Proof 1), which caused me to wonder why it was offered as an olympiad problem. However, the book solution (Proof 2) makes a very elegant shortcut that makes the problem certainly worth looking into.

Both proofs use the fact that \(y=\mbox{cos}(x)\) is monotone decreasing on interval \(\displaystyle (0,\frac{\pi}{2})\):

### Proof 1

\(\displaystyle \begin{align} \mbox{cos}(a) &= \mbox{cos}(\frac{\pi}{2}-b-c) \\ &= \mbox{sin}(b+c) = \mbox{sin}(b)\mbox{cos}(c)+\mbox{sin}(c)\mbox{cos}(b), \end{align} \)

and similarly for \(\mbox{cos}(b)\) and \(\mbox{cos}(c)\). Summing up the three give

\( \begin{align} \mbox{cos}(a)+\mbox{cos}(b)+\mbox{cos}(c) =& \mbox{sin}(a)[\mbox{cos}(b)+\mbox{cos}(c)] \\ &+ \mbox{sin}(b)[\mbox{cos}(c)+\mbox{cos}(a)] \\ &+ \mbox{sin}(c)[\mbox{cos}(a)+\mbox{cos}(b)]. \end{align} \)

Let's focus on one of the terms, say, \(\mbox{sin}(a)[\mbox{cos}(b)+\mbox{cos}(c)]\):

\( \mbox{sin}(a)[\mbox{cos}(b)+\mbox{cos}(c)] = \mbox{sin}(a)\cdot 2\cdot \mbox{cos}(\frac{b+c}{2})\cdot \mbox{cos}(\frac{b-c}{2}). \)

Now, since \(a\gt 0\), \(b+c\lt\frac{\pi}{2}\) and, therefore, \(\frac{b+c}{2}\lt\frac{\pi}{4}\). As we observed at the outset, function \(y=\mbox{cos}(x)\) is monotone decreasing on interval \((0,\frac{\pi}{2})\) such that \(\mbox{cos}(\frac{b+c}{2})\gt \mbox{cos}(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\). For the other factor, we also have \(\mbox{cos}(\frac{b-c}{2})\gt \mbox{cos}(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\), because of the triangle inequality \(|b-c|\le |b|+|c|=b+c\). It follows that

\( \mbox{sin}(a)[\mbox{cos}(b)+\mbox{cos}(c)] \gt \mbox{sin}(a)\cdot 2\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2}}{2} = \mbox{sin}(a). \)

For the other two terms we similarly have \(\mbox{sin}(b)[\mbox{cos}(c)+\mbox{cos}(a)]\gt\mbox{sin}(b)\) and \(\mbox{sin}(c)[\mbox{cos}(a)+\mbox{cos}(b)]\gt\mbox{sin}(c)\). Adding the three up gives the desired inequality.

### Proof 2

Observe that, say, \(a+b\lt\frac{\pi}{2}\) which implies \(a\lt \frac{\pi}{2} - b\), and since \(y=\mbox{cos}(x)\) is monotone decreasing on interval \((0,\frac{\pi}{2})\),

\(\mbox{cos}(a) \gt \mbox{cos}(\frac{\pi}{2} - b) = \mbox{sin}(b).\)

Similarly \(\mbox{cos}(b) \gt \mbox{sin}(c)\) and \(\mbox{cos}(c) \gt \mbox{sin}(a)\). The sum of the three inequalities gives the desired one.

Looking back at the two proofs, it may occur to you that the inequality that has been proved is actually rather weak. Furthermore, as we've seen on the last step of the first proof, \(\mbox{cos}(b)+\mbox{cos}(c)\gt 1\), implying by analogy that \(\mbox{cos}(c)+\mbox{cos}(a)\gt 1\) and \(\mbox{cos}(a)+\mbox{cos}(b)\gt 1\) and so \(\mbox{cos}(a)+\mbox{cos}(b)+\mbox{cos}(c)\gt \frac{3}{2}\). The olympiad inequality would also have been proved had we shown that \(\mbox{sin}(a)+\mbox{sin}(b)+\mbox{sin}(c)\le \frac{3}{2}\), making more precise the notion of the weakness of that inequality.

### Proof 3

The proof is based on the

### Lemma

Let \(a,b,c\) be positive numbers, satisfying \(\displaystyle a+b+c=\frac{\pi}{2}\), prove that

\(\mbox{sin}(a)+\mbox{sin}(b)+\mbox{sin}(c)\le \frac{3}{2}\)

### Proof of Lemma

Instead of an algebraic derivation, I'll base the proof on a geometric insight. First of all note that the graph of \(y = \mbox{sin}(x)\) on interval \((0,\frac{\pi}{2})\) is concave.

Then for three points on the graph that correspond to \(a,b,c\) satisfying \(\displaystyle a+b+c=\frac{\pi}{2}\), the center of gravity is below the graph at the point corresponding to \(\displaystyle\frac{a+b+c}{3}=\frac{\pi}{6}\). It follows that

\(\displaystyle\frac{\mbox{sin}(a)+\mbox{sin}(b)+\mbox{sin}(c)}{3}\le\mbox{sin}(\frac{a+b+c}{3})=\frac{1}{2}\),

with the equality only when \(a=b=c=\frac{\pi}{6}\). Q.E.D.

### Corollary

Assume \(\alpha,\beta,\gamma\) be the angles of an acute triangle, so that each is positive, less than \(\frac{\pi}{2}\) and \(\alpha+\beta+\gamma=\pi\). Then

\(\mbox{cos}(\alpha)+\mbox{cos}(\beta)+\mbox{cos}(\gamma)\le \frac{3}{2}\)

Indeed, consider \(a=\frac{\pi}{2}-\alpha\), \(b=\frac{\pi}{2}-\beta\), \(c=\frac{\pi}{2}-\gamma\), so that \(a+b+c=\frac{\pi}{2}\) and each is positive. From the discussion above,

\(\mbox{sin}(\frac{\pi}{2}-\alpha)+\mbox{sin}(\frac{\pi}{2}-\beta)+\mbox{sin}(\frac{\pi}{2}-\gamma)\le \frac{3}{2}\),

which is equivalent to \(\mbox{cos}(\alpha)+\mbox{cos}(\beta)+\mbox{cos}(\gamma)\le \frac{3}{2}\).

### Proof 4

This proof has been posted below in the comments area. I decided to have it on the page proper for completeness and fairness sake because at the same time two other proofs have been posted at the CutTheKnotMath facebook page which I habitually reproduce at this site.

Let $\alpha = \pi/2 - a$ and so on. So $\alpha + \beta + \gamma = \pi$ and all three angles are in $(0,\pi/2),$ that is, they are the angles of an acute triangle $ABC.$ The inequality is now equivalent to

$\sin\alpha + \sin\beta + \sin\gamma > \cos\alpha + \cos\beta + \cos\gamma.$

Multiplying by the circumdiameter $2R$ and using the law of sines and the fact that $AH = 2Rcos\alpha,$ in which $H$ is the orthocenter, we find that we need to prove that

$BC + AC + BC > AH + BH + CH.$

But $AB$ is longer than the altitude from $A$ of which $AH$ is only a part (here we are using that the triangle is acute), and similarly $BC > BH$ and $CA > CH,$ and we are done.

### Proof 5

The proof is by Leo Giugiuc.

First we prove

### Lemma

Let $\Delta ABC$ be acute angled. Then

$\sin\alpha + \sin\beta + \sin\gamma > \cos\alpha + \cos\beta + \cos\gamma,$

where $\alpha =\angle BAC, $ etc.

Indeed, Let $H$ be the orthocenter of the triangle. Then $BC\gt BH, $AC\gt CH,$ and $AB\gt AH, $ implying

$2R\cdot \sin \alpha + 2R\cdot \sin \beta + 2R\cdot \sin \gamma > 2R\cdot \cos\alpha + 2R\cdot \cos\beta + 2R\cdot \cos\gamma.$

Now back to our problem. Denote $b+c=\alpha,$ $c+a=\beta,$ $a+b=\gamma.$ The three new angles are all in $(0,\pi /2)$ and add up to $\pi.$ From the lemma then, $\sin\alpha + \sin\beta + \sin\gamma > \cos\alpha + \cos\beta + \cos\gamma.$ But

$\sin\alpha=\sin(b+c)=\sin(\pi/2-a)=\cos(a).$

Similarly, $\sin\beta=\cos(b),$ and $\sin\gamma=\cos(c).$ For the same reason $\cos\alpha=\sin(a),$ etc.

### Proof 6

The proof is by Leo Giugiuc.

$a\lt a+b,$ so that $\cos(a)\gt\cos(a+b)=\sin(c).$ Similarly, $\cos(b)\gt\sin(a)$ and $\cos(c)\gt\sin(b).$

### Trigonometry

- What Is Trigonometry?
- Addition and Subtraction Formulas for Sine and Cosine
- The Law of Cosines (Cosine Rule)
- Cosine of 36 degrees
- Tangent of 22.5
^{o}- Proof Wthout Words - Sine and Cosine of 15 Degrees Angle
- Sine, Cosine, and Ptolemy's Theorem
- arctan(1) + arctan(2) + arctan(3) = π
- Trigonometry by Watching
- arctan(1/2) + arctan(1/3) = arctan(1)
- Morley's Miracle
- Napoleon's Theorem
- A Trigonometric Solution to a Difficult Sangaku Problem
- Trigonometric Form of Complex Numbers
- Derivatives of Sine and Cosine
- ΔABC is right iff sin²A + sin²B + sin²C = 2
- Advanced Identities
- Hunting Right Angles
- Point on Bisector in Right Angle
- Trigonometric Identities with Arctangents
- The Concurrency of the Altitudes in a Triangle - Trigonometric Proof
- Butterfly Trigonometry
- Binet's Formula with Cosines
- Another Face and Proof of a Trigonometric Identity
- cos/sin inequality
- On the Intersection of kx and |sin(x)|
- Cevians And Semicircles
- Double and Half Angle Formulas
- A Nice Trig Formula
- Another Golden Ratio in Semicircle
- Leo Giugiuc's Trigonometric Lemma
- Another Property of Points on Incircle
- Much from Little
- The Law of Cosines and the Law of Sines Are Equivalent
- Wonderful Trigonometry In Equilateral Triangle

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