If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$,

Question

If $$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0,$$ prove that $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C).$$

My solution:
From the given, $$\cos^3A+\cos^3B+\cos^3C=3\cos A\cos B\cos C$$

Now, L.H.S$$=\cos3A+\cos3B+\cos3C$$ $$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$ $$=12\cos A \cos B \cos C.$$

My solution ends here. How should I complete?

NOTE: I am not allowed to use complex numbers at my level. So, please help me solve this problem without using complex numbers.

Brute force works: Expand $\cos 3A+\cos 3B+\cos 3C - 3 \cos(A+B+C)$; replace $\sin C$ and $\cos C$ with $-(\sin A+\sin B)$ and $-(\cos A+ \cos B)$, respectively; simplify ... and watch everything magically vanish. — Blue, Feb 13 '16 at 01:39
Here's another insight to help you along. In the step, that is given, write everything as a function of cos(A2). 2Sin^2 A = 1-cos(2A) , 2cos^2(A) = 1 + cos(2A) — Saikat, Feb 13 '16 at 05:01
You aren't "allowed" to use complex numbers? Why would your teacher impose that restriction? — anomaly, Feb 19 '16 at 20:33

score 3 · Accepted Answer · answered Feb 13 '16 at 04:02

3

we have $$sinA+sinB=-sinC$$ and $$cosA+cosB=-cosC$$ squaring and adding both we get

$$2+2cos(A-B)=1$$ that is

$$cos(A-B)=\frac{-1}{2}$$ $\implies$

$$A-B=\frac{2\pi}{3}$$ similarly

$$B-C=\frac{-4\pi}{3}$$ and

$$C-A=\frac{2\pi}{3}$$

Now $$cos3A=cos\left(3\left(B+\frac{2\pi}{3}\right)\right)=cos(3B+2\pi)=cos3B$$

similarly $$cos3C=cos3B$$

Hence

$$cos3A+cos3B+cos3C=3cos3A=3cos(A+A+A)=3cos\left(A+B+\frac{2\pi}{3}+C-\frac{2\pi}{3}\right)=3cos(A+B+C)$$

answered Feb 13 '16 at 04:02

Ekaveera Gouribhatla

14,193

This solution was just wonderful, starting with the first step to write sin A + sin B = -sin C – Saikat Feb 13 '16 at 05:08
Give some tips about developing the intuition. – Saikat Feb 13 '16 at 05:08
My only question is that the process is symmetric. So shouldn't, A-B, b-C, C-A be equal ? – Saikat Feb 13 '16 at 05:11
they cannot be equal because when we add them up you have to get zero – Ekaveera Gouribhatla Feb 13 '16 at 05:14
Why should it be zero ? – Saikat Feb 13 '16 at 05:22
$A-B+B-C+C-A=0$ – Ekaveera Gouribhatla Feb 13 '16 at 05:23
Good observation – Saikat Feb 13 '16 at 05:24
@user230452, Reminds me of http://math.stackexchange.com/questions/1397066/clarification-regarding-a-question – lab bhattacharjee Feb 13 '16 at 10:20
@EkaveeraKumarSharma, $$\cos x=\cos A\implies A=2n\pi\pm A$$ where $n$ is any integer. Why can't $A-B$ be $$-\dfrac{2\pi}3?$$ and why $$B-C$$ has to be $$-\dfrac{4\pi}3$$ ? – lab bhattacharjee Feb 13 '16 at 10:33
you can take like that only thing we have to make sure is $A-B+B-C+C-A=0$...there are infinite solutions and for any solution we get the final result same. – Ekaveera Gouribhatla Feb 13 '16 at 16:47

chenbai · Answer 2 · 2016-02-13T05:19:25.860

1

$\cos (A+B+C)=\cos (A+B) \cos C- \sin (A+B) \sin C \\ = (\cos A \cos B -\sin A \sin B) \cos C -( \sin A \cos B +\cos A \sin B) \sin C \\ =\cos A \cos B \cos C- \sum_{cyc} \sin A \sin B \cos C $

$( \sin A+ \sin B)^2= (-\sin C)^2 , (\cos A +\cos B )^2=(-\cos C)^2 \\ \implies 2+ 2\sin A \sin B +2 \cos A \cos B =1 \\ \implies \sin A \sin B = -\cos A \cos B -\dfrac{1}{2} \\ \implies \sum_{cyc} \sin A \sin B \cos C =\sum_{cyc} (-\cos A \cos B -\dfrac{1}{2})\cos C \\ =-3 \cos A \cos B \cos C -\dfrac{1}{2} (\cos A + \cos B + \cos C)=-3 \cos A \cos B \cos C$

edited Feb 13 '16 at 05:19

answered Feb 13 '16 at 01:51

chenbai

7,681

This is rather confusing to read and leaves some parts out, but this answer works out. – Noble Mushtak Feb 13 '16 at 03:33

score 1 · Answer 3 · answered Feb 19 '16 at 20:25

Trigonometry is not needed here.

You have three unit vectors with sum $0$, which can only happen if the vectors are the sides of an (oriented) equilateral triangle in some order. Geometrically this is the statement that one unit circle centered on another unit circle intersects it at points $\pm 60$ degrees from the center of the first circle.

But in that case $A,B,C$ are (in some order) equal to $\theta, \theta - 120, \theta + 120$ for some angle $\theta$, and we have the stronger result that $3A = 3B = 3C = (A+B+C) = 3\theta$ as angles. Of course the sines and cosines of those angles will also be equal and this implies the given formula.

Tsemo Aristide · Answer 4 · 2016-02-13T02:46:36.637

Hint: Anyway this is a solution with complex numbers. The solution without the complex numbers can be deduced from here, use the formula $e^{ix}=cos(x)+isin(x)$ and compute the real part at it stage of the following method.

$cos(A)+cos(B)+cos(C)=sin(A)+sin(B)+sin(C)=0$ is equivalent to $e^{iA}+e^{iB}+e^{iC}=0$.

$e^{iA}+e^{iB}+e^{iC}=0$ implies $(e^{iA}+e^{iB}+e^{iC})^3=0$.

We use the identity:

$(a+b+c)^3 = a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3bc^2+c^3$

Thus we have:

$e^{3iA}+ 3e^{2iA}e^{iB}+3e^{2iA}e^{iC}+3e^{iA}e^{2iB}+6e^{i(A+B+C)}+3e^{iA}e^{2iC}+e^{3iB}+3e^{2iB}e^{iC}+3e^{iB}e^{2iC}=0$

$=e^{3iA}+e^{3iB}+e^{3iC}+3e^{iA}e^{iB}(e^{iA}+e^{iB})+3e^{iA}e^{iC}(e^{iA}+e^{iC})+3e^{iB}e^{iC}(e^{iB}+e^{iC})+6e^{i(A+B+C)}=0$

We use here $e^{iA}+e^{iB}+e^{iC}=0$ which implies $e^{iA}e^{iB}(e^{iA}+e^{iB})=-e^{iA}e^{iB}e^{iC}$.

We deduce

$e^{3iA}+e^{3iB}+e^{3iC} -3e^{iA}e^{iB}e^{iC}-3e^{iA}e^{iB}e^{iC}-3e^{iA}e^{iB}e^{iC}+6e^{i(A+B+C)}=0$.

$e^{3iA}+e^{3iB}+e^{3iC} = 3(e^{iA}+e^{iB}+e^{iC})$

thus $cos(3A)+cos(3B)+cos(3C)=3(cos(A)+cos(B)+cos(C))$.

"NOTE: I am not allowed to use complex numbers at my level. So, please help me solve this problem without using complex numbers." — Théophile, Feb 13 '16 at 02:34
@Tsemo, An easier proof : http://math.stackexchange.com/questions/404561/if-a-b-c-in-r-are-distinct-then-a3-b3-c33abc-neq-0 — lab bhattacharjee, Feb 13 '16 at 10:28

If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$,

4 Answers4