If $\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$, then $\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $.

Question

If $$\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$$ then how can I show that the sum of cosines of each angle ($x$, $y$, $z$) and sines of each angle sum up to zero? i.e. $$\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $$

I tried:

• Expanded using $\cos(A-B) = \cos A\cos B+\sin A\sin B $, but it did nothing.

After spending one hour to this problem, I thought that there must be a shorter and ideal way.

https://math.stackexchange.com/questions/1397066/clarification-regarding-a-question — lab bhattacharjee, Sep 11 '17 at 10:40
Consider $(\cos x + \cos y + \cos z)^2 + (\sin x + \sin y + \sin z)^2$. — Blue, Sep 11 '17 at 10:47
Note that we have $\sin x+\sin y+\sin z=0=\cos x+\cos y+\cos z$ if and only if the three angles $x,y,z$ are separated by $120^\circ$ pairwise. It would be interesting to see an argument that proves that this follows directly from the cosines of the differences summing to $-\frac32$. — hmakholm left over Monica, Sep 11 '17 at 11:02

Blue · Accepted Answer · 2017-09-11T11:40:43.217

9

By the angle-difference identity you mention, the given equation is equivalent to $$\cos x \cos y + \sin x \sin y + \cos y \cos z + \sin y \sin z + \cos z \cos x + \sin z \sin x = -\frac32 \tag{1}$$ Thus, $$3 + 2 (\cos x \cos y + \cdots ) + 2(\sin x \sin y + \cdots ) = 0 \tag{2}$$

But, $$3 = 1 + 1 + 1 = \left(\cos^2 x + \sin^2 x \right) + \left( \cos^2 y + \sin^2 y \right) + \left( \cos^2 z + \sin^2 z\right) \tag{3}$$

So, (2) becomes $$\begin{align} 0 &= \cos^2 x + \cos^2 y + \cos^2 z + 2 \cos x \cos y + 2 \cos y \cos z + 2\cos z \cos x \\ &+ \sin^2 x + \sin^2 y + \sin^2 z + 2 \sin x \sin y + 2 \sin y \sin z + 2 \sin z \sin x \\ &= \left( \cos x + \cos y + \cos z \right)^2 + \left( \sin x + \sin y + \sin z \right)^2 \end{align} \tag{4}$$

Now, the sum of two squares can be zero only if each square is itself zero, and we are done. $\square$

edited Sep 11 '17 at 11:40

answered Sep 11 '17 at 10:59

Blue

75,673

1

Very nice answer!! – velut luna Sep 11 '17 at 11:00
1

You thought it, the my way – Fghj Sep 11 '17 at 11:04
Can you tell me, how much study, takes to have a brain like you? Please.... – Fghj Sep 11 '17 at 11:11
@RaviPrakash: I appreciate the compliment, but I don't believe that's a question I can answer. :) – Blue Sep 11 '17 at 11:19

Hari Shankar · Answer 2 · 2017-12-01T03:59:27.577

Let $z_1 = \cos x +i \sin x$ etc.

Then $2 \cos (x-y) = \dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}$ etc.

We are given that $\dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}+\dfrac{z_2}{z_3}+\dfrac{z_3}{z_2}+\dfrac{z_3}{z_1}+\dfrac{z_1}{z_3} = -3$

or $\dfrac{z_2+z_3}{z_1}+\dfrac{z_3+z_1}{z_2}+\dfrac{z_1+z_2}{z_3} = -3$

$\Rightarrow \displaystyle \sum_{cyc} \frac{z_2+z_3}{z_1}+1 =0 \Rightarrow (z_1+z_2+z_3)\left(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right) = 0$

Thus $z_1+z_2+z_3 = 0$ or $\displaystyle \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0$

from which the required result follows

score 1 · Answer 3 · answered Sep 11 '17 at 10:48

Let $x-y=\alpha$, $y-z=\beta$. Thus, $z-x=-\alpha-\beta$ and we have $$\cos\alpha+\cos\beta+\cos(\alpha+\beta)+\frac{3}{2}=0$$ or $$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos^2\frac{\alpha+\beta}{2}-1+\frac{3}{2}=0$$ or $$4\cos^2\frac{\alpha+\beta}{2}+4\cos\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}+1=0,$$ which gives $$4\cos^2\frac{\alpha-\beta}{2}-4\geq0$$ or $$\sin\frac{\alpha-\beta}{2}=0.$$ Thus, also $$\left|\cos\frac{\alpha+\beta}{2}\right|=\frac{1}{2}.$$

If $\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$, then $\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $.

3 Answers3

Linked