Proving three angles are equal if they satisfy two conditions

[brotherbobby]

Homework Statement

If $\pmb{A+B+C=\pi}$ and if $\pmb{\sin^2A+\sin^2B+\sin^2C=\sin A\sin B+\sin B\sin C+\sin C\sin A}$, then show that $\pmb{\boxed{A=B=C}}$.

Relevant Equations

1. $\sin x-\sin y=2 \cos\dfrac{x+y}{2} \sin\dfrac{x-y}{2}$
2. $\cos(\pi/2 - \theta)=\sin\theta$

Problem Statement : I copy and paste the statement of the problem directly from the text.

Attempt : I wasn't able to go far into the solution. Below is a rough attempt.

$\begin{equation*} \begin{split} \sin^2A-\sin A\sin B+\sin^2B-\sin B\sin C+\sin^2C-\sin C\sin A & = 0\\ \sin A(\sin A - \sin B)+\sin B (\sin B - \sin C)+\sin C (\sin C - \sin A) & = 0\\ \sin A \cos \dfrac{A+B}{2} \sin \dfrac{A-B}{2}+\sin B \cos \dfrac{B+C}{2} \sin \dfrac{B-C}{2}+\sin C \cos \dfrac{C+A}{2}\sin \dfrac{C-A}{2}&=0\\ \sin A \sin C \sin \dfrac{A-B}{2}+\sin B \sin A \sin \dfrac{B-C}{2}+\sin C \sin B \sin \dfrac{C-A}{2}&=0\\\end{split} \end{equation*}$

I am stuck at the last step.

A hint or clue would be welcome.

 

[PeroK]

I don't see where you used $A + B + C = \pi$? Note that $\sin(\pi - x) = \sin x$.

 

[PeroK]

What about $A = \pi, B = C = 0$?

 

[Vanadium 50]

You have a triangle. Use that fact.

 

[PeroK]

Here's an idea. Try $C = \frac \pi 2$, i.e. $\sin C = 1$. The solution to that is to look at the quantity $(\sin A - \sin B)^2$. Then, try the same for any $C$: i.e. get an expression for $(\sin A - \sin B)^2$.

 

[PeroK]

You have a triangle. Use that fact.

It seems to be more generally true.

 

[SammyS]

You have a triangle. Use that fact.

That seems like a good idea to me, at least if it helps.
If bobby can without this, fine. ... or with it, fine.

 

[Vanadium 50]

It seems to be more generally true.

He has 3 angles that add to 180. How is that more general than a triangle?

 

[PeroK]

He has 3 angles that add to 180. How is that more general than a triangle?

The condition $A + B + C = \pi$ is not needed. If, in the main equality, we replace $\sin A$ by any number $a$, $\sin B$ by any number $b$ and $\sin C$ by any number $c$, then that equality alone implies that $a = b = c$. I found trying to use trig identities a distraction.

 

[PeroK]

For exmple, to give the OP a bit more help. If $c = 2$, then:
$$a^2 + b^2 = ab + 2a + 2b - 4$$And
$$(a-b)^2 = a^2 + b^2 -2ab = 2a + 2b - 4 - ab$$Now use that the LHS is $\ge 0$ to show that $a = b = 2$. Then generalise these steps for any $c$.

It's interesting that trying to see why it doesn't work when $\sin C = 2$ showed that it does work for any numbers!

 

[PeroK]

PPS once we have $\sin A = \sin B = \sin C$, then $A + B + C = \pi$ is one way of constraining the angles so that they must be equal, given the sines are equal. That's where the triangle comes in.

 

[brotherbobby]

I don't see where you used $A + B + C = \pi$? Note that $\sin(\pi - x) = \sin x$.

In step 3, where I used $\cos\dfrac{A+B}{2}=\cos\dfrac{\pi-C}{2}=\sin\dfrac{C}{2}$

 

[PeroK]

In step 3, where I used $\cos\dfrac{A+B}{2}=\cos\dfrac{\pi-C}{2}=\sin\dfrac{C}{2}$

Using trig identities didn't seem to lead anywhere. See the above posts. Sooner or later you have to try something different. I tried $\sin C = 0$ and then $\sin C = 1$ (and that gave me the clue on how to solve it - and to see that it was not directly to do with sines and trig identities).

I never spend too long on something without changing my strategy. For example, if something is hard to prove, then it might not be true. So, you should be checking that out as well.

As above, just take $a, b, c$ instead of $\sin A, \sin B, sin C$ and show that $a = b = c$. Try $c = 1$ to get the algebraic steps needed and then generalise for any $c$.

 

[brotherbobby]

@PeroK , I should confess that I followed little of what you have been saying, but thank you all the same.
I do not want to take any special values for $c=\sin C$, or for any of the other $\sin's$ for that matter. The solution I am looking for should follow straight to the equality $A=B=C$.

Look at post #3 above, from you.

What about $A = \pi, B = C = 0$?

In that case, the given conditional $\sin^2A+\sin^2B+\sin^2C=\sin A\sin B+\sin B\sin C+\sin C\sin A$ would not be true, for the L.H.S = 1 while the R.H.S. = 0. This conditional is a given, so it cannot be taken to be something else.

 

[brotherbobby]

Using trig identities didn't seem to lead anywhere. See the above posts. Sooner or later you have to try something different. I tried $\sin C = 0$ and then $\sin C = 1$ (and that gave me the clue on how to solve it - and to see that it was not directly to do with sines and trig identities).

I never spend too long on something without changing my strategy. For example, if something is hard to prove, then it might not be true. So, you should be checking that out as well.

As above, just take $a, b, c$ instead of $\sin A, \sin B, sin C$ and show that $a = b = c$. Try $c = 1$ to get the algebraic steps needed and then generalise for any $c$.

Ok let me think along your lines. It is not the ideal way to solve this problem, given that it is from a text on elementary trigonometry. But if nothing else works out, it is better than nothing.

 

[PeroK]

I do not want to take any special values for $c=\sin C$, or for any of the other $\sin's$ for that matter.

Why not? That's how you gain insight into a problem.

The solution I am looking for should follow straight to the equality $A=B=C$.

There are two modes of mathematical thinking: ("machine mode") working inside the system, where you strictly apply mathematical rules; and, "intelligence mode", working outside the system, where you ask yourself "is this true?", "why is this true?", "what if I try this?" ...

You're trying to do mathematics almost entirely in machine mode and not using intelligence mode. That's partly why you go round in circles. Whereas, I jumped out of the system, worked in intelligence mode and then saw what machine rules to apply.

Using your intelligence is the difference between solving these problems and being able to move on and endlessly applying mathematical rules until, perhaps eventually, you stumble blindly on a solution.

 

[brotherbobby]

Why not? That's how you gain insight into a problem.

There are two modes of mathematical thinking: ("machine mode") working inside the system, where you strictly apply mathematical rules; and, "intelligence mode", working outside the system, where you ask yourself "is this true?", "why is this true?", "what if I try this?" ...

You're trying to do mathematics almost entirely in machine mode and not using intelligence mode. That's partly why you go round in circles. Whereas, I jumped out of the system, worked in intelligence mode and then saw what machine rules to apply.

Using your intelligence is the difference between solving these problems and being able to move on and endlessly applying mathematical rules until, perhaps eventually, you stumble blindly on a solution.

Yes I understand. Very good point. I suppose it is how someone's been trained. The country I am from, we train our children in the machine mode. I suppose both the machine and intelligence mode are necessary to be a good math student.

 

[brotherbobby]

Why not? That's how you gain insight into a problem.

There are two modes of mathematical thinking: ("machine mode") working inside the system, where you strictly apply mathematical rules; and, "intelligence mode", working outside the system, where you ask yourself "is this true?", "why is this true?", "what if I try this?" ...

You're trying to do mathematics almost entirely in machine mode and not using intelligence mode. That's partly why you go round in circles. Whereas, I jumped out of the system, worked in intelligence mode and then saw what machine rules to apply.

Using your intelligence is the difference between solving these problems and being able to move on and endlessly applying mathematical rules until, perhaps eventually, you stumble blindly on a solution.

Yes I understand. Very good point. I suppose it is how someone's been trained. The country I am from, we train our children in the machine mode. I suppose both the machine and intelligence mode are necessary to be a good math student.

 

[brotherbobby]

Ok I have solved the problem using @PeroK 's technique. I provide the details below.

Problem statement : Given $A+B+C=\pi$ ($\leftarrow$we'd find this not necessary) and $\sin^2A+\sin^2B+\sin^2C=\sin A\sin B+\sin B\sin C+\sin C\sin A$, show that angles $\boxed{A=B=C}$.

Attempt : Let us take $\sin A =a, \sin B=b, \sin C=c$. This makes our given (2nd) identity appear thus
\begin{equation*}
\begin{split}
a^2+b^2+c^2 &= ab+bc+ca\\
\Rightarrow a^2+b^2+c^2-ab-bc-ca &= 0 \\
\Rightarrow \dfrac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]&=0 ^{\large{\pmb{*}}}\\
\Rightarrow a&=b=c\\
\Rightarrow \sin A & = \sin B=\sin C\\
\Rightarrow \pmb{A}&=\pmb{B=C}
\end{split}
\end{equation*}
$^{\large{\pmb{*}}}$ can be shown via elementary algebra

(In retrospect, I suppose we do need the fact that the angles $A+B+C=\pi$. This would constrain the equality of their sine's to the principal values of the angles)