Proving a Trigonometric equality

[anemone]

Hi MHB,

I have found this problem quite interesting to me and hence I have spent some time on it but all of my attempts to prove it went down the drain.

I have no choice but posting it here, hoping to gain some insight from the members of the forum on how to prove this problem.

Thanks in advance.

Problem:

Suppose that real numbers $a, b, c$ satisfy

$\dfrac{\cos a+\cos b+\cos c}{\cos(a+b+c)}=\dfrac{\sin a+\sin b+\sin c}{\sin(a+b+c)}=p$

Prove that $\cos(a+b)+\cos(b+c)+\cos(a+c)=p$

 

[mente oscura]

Hi MHB,

I have found this problem quite interesting to me and hence I have spent some time on it but all of my attempts to prove it went down the drain.

I have no choice but posting it here, hoping to gain some insight from the members of the forum on how to prove this problem.

Thanks in advance.

Problem:

Suppose that real numbers $a, b, c$ satisfy

$\dfrac{\cos a+\cos b+\cos c}{\cos(a+b+c)}=\dfrac{\sin a+\sin b+\sin c}{\sin(a+b+c)}=p$

Prove that $\cos(a+b)+\cos(b+c)+\cos(a+c)=p$

Hello.

But the question is: always is fulfilled?:

\cos(a+b)+\cos(b+c)+\cos(a+c)=p

, or what can there be a solution?.

I have come to a conclusion in favour of the second question.

Regards.

 

[Opalg]

Suppose that real numbers $a, b, c$ satisfy

$\dfrac{\cos a+\cos b+\cos c}{\cos(a+b+c)}=\dfrac{\sin a+\sin b+\sin c}{\sin(a+b+c)}=p$

Prove that $\cos(a+b)+\cos(b+c)+\cos(a+c)=p$

I will use a trick that I learned from this forum: if $\dfrac xy = \dfrac zw = p$ then also $\dfrac{\lambda x + \mu z}{\lambda y + \mu w} = p$ for any choice of $\lambda$ and $\mu$.

Given $\dfrac{\cos a+\cos b+\cos c}{\cos(a+b+c)}=\dfrac{\sin a+\sin b+\sin c}{\sin(a+b+c)}=p$, apply that trick with $\lambda = \cos c$ and $\mu = \sin c$, to get $$p = \frac{\cos c(\cos a+\cos b+\cos c) + \sin c(\sin a+\sin b+\sin c)}{\cos c\cos(a+b+c) + \sin c\sin(a+b+c)} = \frac{\cos(c-a) + \cos(c-b) + 1}{\cos(a+b)}.$$ Similarly, using $\lambda = \sin c$ and $\mu = -\cos c$, you get $$p = \frac{\sin c(\cos a+\cos b+\cos c) - \cos c(\sin a+\sin b+\sin c)}{\sin c\cos(a+b+c) - \cos c\sin(a+b+c)} = \frac{\sin(c-a) + \sin(c-b)}{-\sin(a+b)}.$$ Thus $$\frac{\cos(c-a) + \cos(c-b) + 1}{\cos(a+b)} = \frac{\sin(c-a) + \sin(c-b)}{-\sin(a+b)} = p.$$ Now apply the trick again, to that expression, this time with $\lambda = \cos(a+b)$ and $\mu = -\sin(a+b)$: $$\begin{aligned} p &= \frac{\cos(a+b)\bigl(\cos(c-a) + \cos(c-b) + 1\bigr) - \sin(a+b)\bigl(\sin(c-a) + \sin(c-b)\bigr)}{\cos^2(a+b) + \sin^2(a+b)} \\ &= \frac{\cos(b+c) + \cos(a+c) + \cos(a+b)}{1}. \end{aligned}$$ Therefore $\cos(b+c) + \cos(a+c) + \cos(a+b) = p$.

 

[mente oscura]

I will use a trick that I learned from this forum: if $\dfrac xy = \dfrac zw = p$ then also $\dfrac{\lambda x + \mu z}{\lambda y + \mu w} = p$ for any choice of $\lambda$ and $\mu$.

Given $\dfrac{\cos a+\cos b+\cos c}{\cos(a+b+c)}=\dfrac{\sin a+\sin b+\sin c}{\sin(a+b+c)}=p$, apply that trick with $\lambda = \cos c$ and $\mu = \sin c$, to get $$p = \frac{\cos c(\cos a+\cos b+\cos c) + \sin c(\sin a+\sin b+\sin c)}{\cos c\cos(a+b+c) + \sin c\sin(a+b+c)} = \frac{\cos(c-a) + \cos(c-b) + 1}{\cos(a+b)}.$$ Similarly, using $\lambda = \sin c$ and $\mu = -\cos c$, you get $$p = \frac{\sin c(\cos a+\cos b+\cos c) - \cos c(\sin a+\sin b+\sin c)}{\sin c\cos(a+b+c) - \cos c\sin(a+b+c)} = \frac{\sin(c-a) + \sin(c-b)}{-\sin(a+b)}.$$ Thus $$\frac{\cos(c-a) + \cos(c-b) + 1}{\cos(a+b)} = \frac{\sin(c-a) + \sin(c-b)}{-\sin(a+b)} = p.$$ Now apply the trick again, to that expression, this time with $\lambda = \cos(a+b)$ and $\mu = -\sin(a+b)$: $$ p = \frac{\cos(a+b)\bigl(\cos(c-a) + \cos(c-b) + 1\bigr) - \sin(a+b)\bigl(\sin(c-a) + \sin(c-b)\bigr)}{\cos^2(a+b) + \sin^2(a+b)} = \frac{\cos(b+c) + \cos(a+c) + \cos(a+b)}{1}.$$ Therefore $\cos(b+c) + \cos(a+c) + \cos(a+b) = p$.

Hello.

Yes, but:

cos \ a+cos \ b+cos \ c=p \ cos \ (a+b+c)

sin \ a+sin \ b+sin \ c=p \ sin \ (a+b+c)

p^2 \ cos^2(a+b+c)=cos^2a+cos^2b+cos^2c+2cos a \ cos b+2cos a \ cos c+2cos b \ cos c, (*)

p^2 \ sin^2(a+b+c)=sin^2a+sin^2b+sin^2c+2sin a \ sin b+2sin a \ sin c+2sin b \ sin c, (**)

I'm add (*) and (**)

p^2=3+2cos a \ cos b+2cos a \ cos c+2cos b \ cos c+2sin a \ sin b+2sin a \ sin c+2sin b \ sin c

p^2=3+4\cos a \cos b+4\cos a \cos c+4\cos b \cos c+2\cos (a+b)+2\cos (a+c)+2\cos (b+c)We substitute, to see if it can meet that:

\cos(b+c) + \cos(a+c) + \cos(a+b) = p

2p=2 \cos(b+c) +2 \cos(a+c) +2 \cos(a+b)

p^2-2p-4(\cos a \cos b+\cos a \cos c+\cos b \cos c)-3=0

p=-1 \pm \sqrt{1+\cos a \cos b+\cos a \cos c+\cos b \cos c}

Could we discard the negative outcome?

Regards.

 

[anemone]

I will use a trick that I learned from this forum: if $\dfrac xy = \dfrac zw = p$ then also $\dfrac{\lambda x + \mu z}{\lambda y + \mu w} = p$ for any choice of $\lambda$ and $\mu$.

Given $\dfrac{\cos a+\cos b+\cos c}{\cos(a+b+c)}=\dfrac{\sin a+\sin b+\sin c}{\sin(a+b+c)}=p$, apply that trick with $\lambda = \cos c$ and $\mu = \sin c$, to get $$p = \frac{\cos c(\cos a+\cos b+\cos c) + \sin c(\sin a+\sin b+\sin c)}{\cos c\cos(a+b+c) + \sin c\sin(a+b+c)} = \frac{\cos(c-a) + \cos(c-b) + 1}{\cos(a+b)}.$$ Similarly, using $\lambda = \sin c$ and $\mu = -\cos c$, you get $$p = \frac{\sin c(\cos a+\cos b+\cos c) - \cos c(\sin a+\sin b+\sin c)}{\sin c\cos(a+b+c) - \cos c\sin(a+b+c)} = \frac{\sin(c-a) + \sin(c-b)}{-\sin(a+b)}.$$ Thus $$\frac{\cos(c-a) + \cos(c-b) + 1}{\cos(a+b)} = \frac{\sin(c-a) + \sin(c-b)}{-\sin(a+b)} = p.$$ Now apply the trick again, to that expression, this time with $\lambda = \cos(a+b)$ and $\mu = -\sin(a+b)$: $$\begin{aligned} p &= \frac{\cos(a+b)\bigl(\cos(c-a) + \cos(c-b) + 1\bigr) - \sin(a+b)\bigl(\sin(c-a) + \sin(c-b)\bigr)}{\cos^2(a+b) + \sin^2(a+b)} \\ &= \frac{\cos(b+c) + \cos(a+c) + \cos(a+b)}{1}. \end{aligned}$$ Therefore $\cos(b+c) + \cos(a+c) + \cos(a+b) = p$.

Hi Opalg,

I must say a big thank you to you, who has constantly helped me out to understand how to use some nice tricks to solve some challenging mathematics problems and words cannot adequately express my deepest gratitude to you for the remarkable helps that you have shown to me.(Sun)(heart)

Also, even if I'm aware of the technique to use the following trick, it would take me some time to figure out the proper $\lambda$ and $\mu$ to be used!

($\dfrac xy = \dfrac zw = p$ then also $\dfrac{\lambda x + \mu z}{\lambda y + \mu w} = p$ for any choice of $\lambda$ and $\mu$.)

Hello.

p^2 \ cos^2(a+b+c)=cos^2a+cos^2b+cos^2c+2cos a \ cos b+2cos a \ cos c+2cos b \ cos c, (*)

p^2 \ sin^2(a+b+c)=sin^2a+sin^2b+sin^2c+2sin a \ sin b+2sin a \ sin c+2sin b \ sin c, (**)

I'm add (*) and (**)

p^2=3+2cos a \ cos b+2cos a \ cos c+2cos b \ cos c+2sin a \ sin b+2sin a \ sin c+2sin b \ sin c

p^2=3+4\cos a \cos b+4\cos a \cos c+4\cos b \cos c+2\cos (a+b)+2\cos (a+c)+2\cos (b+c)

Shouldn't the above be p^2=3+4\sin a \sin b+4\sin a \sin c+4\sin b \sin c+2\cos (a+b)+2\cos (a+c)+2\cos (b+c)?

 

[mente oscura]

Hello.

I'm add (*) and (**)

p^2=3+2cos a \ cos b+2cos a \ cos c+2cos b \ cos c+2sin a \ sin b+2sin a \ sin c+2sin b \ sin c

p^2=3+4\cos a \cos b+4\cos a \cos c+4\cos b \cos c+2\cos (a+b)+2\cos (a+c)+2\cos (b+c)We substitute, to see if it can meet that:

\cos(b+c) + \cos(a+c) + \cos(a+b) = p

2p=2 \cos(b+c) +2 \cos(a+c) +2 \cos(a+b)

p^2-2p-4(\cos a \cos b+\cos a \cos c+\cos b \cos c)-3=0

p=-1 \pm \sqrt{1+\cos a \cos b+\cos a \cos c+\cos b \cos c}

Could we discard the negative outcome?

Regards.

Hello.

I'm sorry. I am wrong in the transcription of any sign. I this correct:

p^2=3+2cos a \ cos b+2cos a \ cos c+2cos b \ cos c+2sin a \ sin b+2sin a \ sin c+2sin b \ sin c

By be:

\cos (a+b)= \cos a \cos b-\sin a \sin b

2 \sin a \sin b=2 \cos a \cos b-2 \cos (a+b)

This is similar with \cos (a+c) \ and \ \cos (b+c)

It would be:

p^2=3+4 \cos a \cos b+4 \cos a \cos c+4 \cos b \cos c-2 \cos (a+b)-2 \cos (a+c)-2 \cos (b+c)

p^2=3+4 \cos a \cos b+4 \cos a \cos c+4 \cos b \cos c-2 p

p^2+2p-4(\cos a \cos b+\cos a \cos c+\cos b \cos c)-3=0p=-1 \pm \sqrt{1+\cos a \cos b+\cos a \cos c+\cos b \cos c}Regards.