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The problem is as follows:

Using:

$A-B+C=360^\circ$

Find as product the expression from below:

$\sin 2A-\sin 2B+\sin 2C$

The alternatives given by my book are as follows:

$\begin{array}{ll} 1.&2\sin A\sin B\sin C\\ 2.&4\cos A \cos B \cos C\\ 3.&4\sin A \sin B \sin C\\ 4.&-2\sin A \cos B \sin C\\ \end{array}$

Please take notice that in order to avoid any sort of duplication with this question and other similars. Please note that the signs are not the same. And the angle condition is not the same.

I'm rephrasing my question as an individual situation based on my strategy on how to solve this problem provided that I landed in the right answer. Thus the sort of answer which I'm intending to get is how to solve or prove this without just many steps and more logically and less prone to get errors in algebraic manipulation.

What I've attempted to so, was a similar strategy as seen in the Cagnoli's identity.

Thus:

$A-B+c=2\pi$

Then:

$\sin (A-B+C)=\sin 2\pi=0$

$\sin (A-B)\cos C+\cos (A-B)\sin C=0$

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C + \sin A \sin B \sin C=0$

Then I could spot:

$-\sin A \sin B \sin C=\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

Now the thing is how to get what it is requested:

Then I thought:

On the right side of the equation I'm getting:

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

$\frac{1}{2}(\sin (A+B)+\sin (A-B))\cos C - \frac{1}{2}\cos C (\sin (A+B)-\sin (A-B))+\frac{1}{2}(\cos (A+B)+\cos (A-B))\sin C$

Multiplying by $2$ on the whole expression and replacing from:

$A-B=2\pi-C$

$(\sin (A+B)+\sin (2\pi-C))\cos C - \cos C (\sin (A+B)-\sin (2\pi-C))+(\cos (A+B)+\cos (2\pi-C))\sin C = -2\sin A \sin B \sin C$

$(\sin (A+B)-\sin (C))\cos C - \cos C (\sin (A+B)+\sin (C))+(\cos (A+B)+\cos (C))\sin C$

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B)-\frac{1}{2}\sin 2C +\sin C \cos (A+B) + \frac{1}{2}\sin 2C$

Simplifying:

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B) +\sin C \cos (A+B)$

Multiplying by $2$:

$2\cos C \sin (A+B)-\sin 2C- 2\cos C \sin (A+B) +2\sin C \cos (A+B)=-4\sin A \sin B \sin C$

Then product to sum:

$\sin (A+B+C)-\sin (C-A-B)-\sin 2C - \sin (A+B+C)+\sin (C-A-B)+\sin (A+B+C)+\sin (C-A-B)$

Luckily there are some terms which can be simplified:

$-\sin 2C +\sin (A+B+C)+\sin (C-A-B)$

Then I guess that what is left to replace is : $A-B+C=2\pi$

$-\sin 2C + \sin (2\pi+2B)+\sin (2\pi-2A)$

Then this becomes into:

$-\sin 2C + sin 2B -\sin 2A= -4\sin A \sin B \sin C$

Since it is mentioned that it requests to obtain the identity from the mentioned expression, this translates into:

Multiplying by $-1$:

$\sin 2A-\sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

Therefore that correspond with the option 3. And I believe this must be the answer. If my algebra is correct. I wonder, if it does exist a less painful way to get the same result without just relying in too many lines and perhaps. Did I overlooked something?. It would help me a lot if someone could help me with this. As my methodology seems too long and I had to do several trials to get this right.

Chris Steinbeck Bell
  • 4,175

3 Answers3

1

Solving by algebra and identities is absolutely the hardest possible way to proceed.

Instead, pick values of $A$, $B$, and $C$ that satisfy the relation $A - B + C = 360^\circ$ and then see which of the potential answers are eliminated. (It can take a few rounds of choices to become familiar with how certain choices make certain terms be zero or $\pm 1$, and I haven't shown these exploratory steps. The choices I've made below were my first choices and come from much prior practice.)

  • For instance, take $A = 360^\circ$, and $B = C = 0^\circ$. We find $\sin 2A - \sin 2B + \sin 2C = 0$. Any product with a $\sin A$, a $\sin B$, or a $\sin C$ is also zero, so we have probably eliminated potential answer 2. We compute $4 \cos A \cos B \cos C = 4 \cdot 1 \cdot 1 \cdot 1 \neq 0$, so we have eliminated potential answer 2.

All the other answers have $\sin A$ instead of $\cos A$, so it's time to pick values for $A$ that have $\sin A \neq 0$.

  • Now try $A = C = 90^\circ$, $B = -180^\circ$. Again, $\sin 2A - \sin 2B + \sin 2C = 0$. Here, potential answer 4 gives $-2 \sin A \cos B \sin C = -2 \cdot 1 \cdot -1 \cdot 1 \neq 0$, so we have eliminated potential answer 4.

The surviving potential answers differ only by a factor of $2$, so we want all three of $\sin A$, $\sin B$, and $\sin C$ different from 0.

  • Take $A = 270^\circ$, $B = 30^\circ$ and $C = 120^\circ$. We find $\sin 2A - \sin 2B + \sin 2C = 0 - \sqrt{3}/2 -\sqrt{3}/2 = -\sqrt{3}$. Then, $\sin A \sin B \sin C = -1 \cdot 1/2 \cdot \sqrt{3}/2 = -\sqrt{3}/4$, eliminating potential answer 1.

So we know that potential answer 3 is the correct answer.

Eric Towers
  • 67,037
1

Note

\begin{align} & \sin 2A-\sin 2B+\sin 2C\\ = & 2\sin A \cos A -2\sin(B-C) \cos(B+C)\\ = &2\sin A \sin(2\pi +B- C) -2\sin (A-2\pi ) \cos(B+ C )\\ = & 2\sin A\sin (B-C) -2\sin A \cos(B+ C )\\ = & 2\sin A[ \cos ( B-C) -\cos(B+C)] \\ = & 4\sin A \sin B\sin C \\ \end{align}

Quanto
  • 97,352
0

Let's do some transformation first to turn this into a symmetric problem. Set A=a, B=-b, C= c

then $a+b+c=360^\circ$

$sin (a+b)=-sinc$

$\cos (a+b)=cosc$

$sin2a+sin2b+sin2c$

$= 2sin(a+b)*cos(a-b)+sin2c$ $= -2sin c* cos(a-b)+2sinc*cosc$

$=2sinc* (cosc-cos(a-b))$ $= 2sinc*(cos(a+b)-cos(a-b))$ $=2sinc*(-2sina*sinb)$ $=-4sina*sinb*sinc$

Replace a,b,c with A,B,C, we have $\sin 2A-\sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

Star Bright
  • 2,338