Hard contest type trigonometry proof

Question

Suppose that real numbers $x, y, z$ satisfy: $$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)} = \frac{\sin x + \sin y + \sin z}{\sin (x + y + z )} = p$$

Then prove that: $$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$

I am not even getting where to start? Please help.

tried expanding $\cos (x + y) + \cos (y + z ) + \cos (x + z)$? — , Nov 18 '13 at 10:55
even if that was not helpful to you,if you can write that here some one can extend that to desired result... — , Nov 18 '13 at 10:59
@PraphullaKoushik I have expanded that and nothing else, which is very obvious to do. — Sawarnik, Nov 18 '13 at 11:03
Some observations, not sure if helpful $$p^2\cos^2(x+y+z)=(\sum \cos x)^2=\sum\cos^2x+2\sum\cos x\cos y$$
$$p^2\sin^2(x+y+z)=(\sum \sin x)^2=\sum\sin^2x+2\sum\sin x\sin y$$

Adding we get $$p^2=3+2\sum\cos(x-y)$$

On subtraction, $$p^2\cos(2x+2y+2z)=\sum\cos2x+2\sum \cos(x+y)$$ — lab bhattacharjee, Nov 18 '13 at 11:23
@Sawarnik, would you mind sharing the source of the Hard (the hardest I have met with recently)problem? — lab bhattacharjee, Nov 18 '13 at 11:26
@labbhattacharjee I have PDF called Trigonometry Problems, I don't remember the source but it contains 53 equally hard problems. — Sawarnik, Nov 18 '13 at 11:53
@Sawarnik,can you link your Trigonometry problems PDF? Thank you — math110, Nov 18 '13 at 11:55
@math110 http://ohkawa.cc.it-hiroshima.ac.jp/AoPS.pdf/Trigonometry%20Problems%20-%20Amir%20Hossein%20Parvardi.pdf — Sawarnik, Nov 18 '13 at 11:58
@Sawarnik, thanks a ton for the link. Some of the problems are already logged here:) — lab bhattacharjee, Nov 18 '13 at 12:06
Does it mind if I post several more questions that I could not solve [infact most of the questions] as separate questions? — Sawarnik, Nov 18 '13 at 12:27
@Sawarnik, try to post independent problems in as separate questions — lab bhattacharjee, Nov 18 '13 at 14:35

score 15 · Accepted Answer · answered Nov 18 '13 at 11:28

15

note $$\cos{(x+y)}=\cos{[(x+y+z)-z]}=\cos{(x+y+z)}\cos{z}+\sin{(x+y+z)}\sin{z}$$ and $$\cos{(y+z)}=\cos{(x+y+z)}\cos{x}+\sin{(x+y+z)}\sin{x}$$ $$\cos{(z+x)}=\cos{(x+y+z)}\cos{y}+\sin{(x+y+z)}\sin{y}$$ add this three \begin{align*} &\cos{(x+y)}+\cos{(y+z)}+\cos{(x+z)}\\ &=(\cos{x}+\cos{y}+\cos{z})\cos{(x+y+z)}+(\sin{x}+\sin{y}+\sin{z})\sin{(x+y+z)}\\ &=p\cos^2{(x+y+z)}+p\sin^2{(x+y+z)}\\ &=p \end{align*}

answered Nov 18 '13 at 11:28

math110

93,304

1

nice solution, this also solves another problem : $$\sum\sin(x+y)=0$$ – lab bhattacharjee Nov 18 '13 at 11:37
1

Wow, this is really elegant! – Sawarnik Nov 18 '13 at 11:55

score 6 · Answer 2 · answered Nov 18 '13 at 12:00

I prefer @math110's solution, but here's a brute force method using complex exponentials, with $$\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \qquad \sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$ We define $$a := e^{ix} \qquad b := e^{iy} \qquad c := e^{iz}$$ so that $$p = \frac{\cos x + \cos y + \cos z}{\cos(x+y+z)} \implies p(a^2b^2c^2 +1) = abc (a+b+c) + bc + ca + ab \qquad (1)$$ $$p = \frac{\sin x + \sin y + \sin z}{\sin(x+y+z)} \implies p(a^2b^2c^2 - 1 ) = abc (a+b+c) - bc - ca - ab \qquad (2)$$

Thus, from $(1)-(2)$ and $(1)+(2)$, we have $$p = bc + ca + ab \qquad\qquad p = \frac{a+b+c}{abc} = \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$$ whereupon $$2p = bc+\frac{1}{bc}\;+\;ca+\frac{1}{ca}\;+\;ab+\frac{1}{ab} = 2\left( \cos(y+z)+\cos(z+x)+\cos(x+y)\right)$$

score 3 · Answer 3 · answered Jan 26 '17 at 08:52

Since \begin{align*} p \cos(x+y+z) &= \cos x + \cos y + \cos z \\ p \sin(x+y+z) &= \sin x + \sin y +\sin z \end{align*} \begin{align*} p e^{i(x+y+z)} & = e^{ix} + e^{iy} + e^{iz} \end{align*} Multiplying throughout by $e^{-i(x+y+z)}$, we get \begin{align*} p = e^{-i(y+z)} + e^{-i(z+x)} + e^{-i(x+y)} \end{align*} Since $p$ is real, equating the real parts, we get $$\cos(y+z) + \cos(z+x) + \cos(x+y) = p $$

Hard contest type trigonometry proof

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