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I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:

Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).

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    $\begingroup$ Your intuition that the laws of sine and cosine are in play is doubtless valid. Since the present form does not seem similar to either of those, try clearing the denominators, either by dividing through or by multiplying/factoring them out. Note that the correspondence of (say) angle $\alpha$ opposite side $a$, etc. must come into play and should be stated. $\endgroup$ – hardmath Jun 15 '15 at 23:20
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    $\begingroup$ By the way, has anyone ever used the law of sines or cosines outside of high school level mathematics? As a trained mathematician, I can't remember ever seeing it during university or professional years. $\endgroup$ – Simon S Jun 16 '15 at 13:54
  • $\begingroup$ I wonder if there is a more transparent geometric demonstration (I hesitate to call it a proof). Each of the fractions can be interpreted as a kind of slope of the line from a vertex to the midpoint of the opposite side, relative to the altitude from that vertex to the opposite side (and taking orientation into account). The sum of the fractions being zero then has a more or less natural interpretation. $\endgroup$ – Brian Tung Jun 16 '15 at 16:53
  • $\begingroup$ Equivalently: Draw lines from a vertex to the orthocenter, and to the centroid. Determine the tangent of that angle (where if the orthocenter is "counterclockwise" from the centroid, relative to the vertex, the angle is taken to be negative). Then the sum of those tangents, for all three vertices, is equal to zero. (That's not a proof, it's a restatement.) $\endgroup$ – Brian Tung Jun 16 '15 at 16:56
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The Law of Cosines is equivalent to (and is often proven via) the statements $$a = b \cos\gamma + c\cos\beta \qquad b = c \cos\alpha + a \cos\gamma \qquad c = a \cos\beta + b \cos\alpha$$ so, your sum becomes $$\frac{c \cos\alpha - a \cos\gamma}{a\sin\gamma} + \frac{a \cos\beta - b \cos\alpha}{b\sin\alpha} + \frac{b \cos\gamma - c \cos\beta}{c\sin\beta}$$

Further, the Law of Sines allows us to write $$a = d \sin\alpha \qquad b = d\sin\beta \qquad c = d \sin\gamma$$ where $d$ is the triangle's circumdiameter. This gives

$$\frac{\sin\gamma \cos\alpha - \sin\alpha \cos\gamma}{\sin\alpha\sin\gamma} + \frac{\sin\alpha \cos\beta - \sin\beta \cos\alpha}{\sin\beta\sin\alpha} + \frac{\sin\beta \cos\gamma - \sin\gamma \cos\beta}{\sin\gamma\sin\beta}$$ $$= \left( \cot\alpha - \cot\gamma \right) + \left( \cot\beta - \cot\alpha \right) + \left( \cot\gamma - \cot\beta \right ) = 0 $$

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$\displaystyle \sum_{cyclic} \dfrac{b-2a\cos \gamma}{a\sin \gamma}=\displaystyle \sum_{cyclic} \dfrac{b^2-2ab\cos \gamma}{ab\sin \gamma}= \dfrac{1}{2S}\displaystyle \sum_{cyclic} (b^2-2ab\cos \gamma)=0$ by the Cosine Law.

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    $\begingroup$ It seems to me that this way of solving is surpassing the high school mathematics. Could you please give the another, more trivial way of solving this? $\endgroup$ – GreatDuke Jun 15 '15 at 23:25
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    $\begingroup$ It is only notationally that the answer is not in line with high school math. Use the idea. For example your first term is $\frac{b^2-2ab\cos \gamma}{ab\sin\gamma}$. The bottom is twice the area of the triangle, in the notation above $2S$. And by the Cosine Law the top is $c^2-a^2$. So the first term is $\frac{c^2-a^2}{2S}$. Work out the second term and the third term similarly, and add. $\endgroup$ – André Nicolas Jun 16 '15 at 0:02
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Consider the first summand, to begin with: $$ C=\frac{b-2a\cos\gamma}{a\sin\gamma} $$ The law of cosines tells you that $$ c^2=a^2+b^2-2ab\cos\gamma $$ so we can write $$ b-2a\cos\gamma=\frac{c^2-a^2}{b} $$ and so we have $$ C=\frac{c^2-a^2}{ab\sin\gamma} $$ The law of sines is $$ \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R $$ where $R$ is the radius of the circumscribed circle. In particular, $$ \sin\gamma=\frac{c}{2R} $$ and so we have $$ C=\frac{2R(c^2-a^2)}{abc} $$ Similarly, $$ A=\frac{c-2b\cos\alpha}{b\sin\alpha}=\frac{a^2-b^2}{abc}\\ B=\frac{a-2c\cos\beta}{c\sin\beta}=\frac{b^2-c^2}{abc} $$ and $$ C+A+B=\frac{c^2-a^2+a^2-b^2+b^2-c^2}{abc}=0 $$

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$\dfrac{b - 2a\cos{\gamma} }{a\sin{\gamma}}=\dfrac{\sin{\beta}}{\sin{\alpha}\sin{\gamma}}-\dfrac{2\cos{\gamma}}{\sin{\gamma}}$

$\sum\dfrac{b - 2a\cos{\gamma} }{a\sin{\gamma}}=\sum \dfrac{\sin^2{\alpha}}{\sin{\alpha}\sin{\beta}\sin{\gamma}}-\sum \dfrac{2\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}\sin{\gamma}}= \dfrac{\sum\sin{\alpha}(\sin{\alpha}-2\cos{\gamma}\sin{\beta})}{\sin{\alpha}\sin{\beta}\sin{\gamma}}$

$\sin{\alpha}(\sin{\alpha}-2\cos{\gamma}\sin{\beta})=\sin{(\beta+\gamma)}(\sin{(\beta+\gamma)}-2\cos{\gamma}\sin{\beta})=\sin{(\beta+\gamma)}(\cos{\beta}\sin{\gamma}-\cos{\gamma}\sin{\beta})=\sin{(\beta+\gamma)}\sin{(\gamma-\beta)}=\dfrac{1}{2}(\cos{2\beta}-\cos{2\gamma}) \implies \sum\sin{\alpha}(\sin{\alpha}-2\cos{\gamma}\sin{\beta})=\dfrac{1}{2}\sum (\cos{2\beta}-\cos{2\gamma}) =0$

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From Sine Rule in the triangle, we have $$\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}=\frac{\sin\gamma}{c}=k \space (\text{let any arbitrary constant})$$ Thus by substituting the values of $a=k\sin \alpha$, $b=k\sin \beta$ & $c=k\sin \gamma$ in the given expression, we get $$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\cos\alpha}{b\sin\alpha}+\frac{a-2c\cos\beta}{c\sin\beta}$$ $$=\frac{k\sin\beta-2k\sin\alpha\cos\gamma}{k\sin\alpha\sin\gamma}+\frac{k\sin\gamma-2k\sin\beta\cos\alpha}{k\sin\beta\sin\alpha}+\frac{k\sin\alpha-2k\sin\gamma\cos\beta}{k\sin\gamma\sin\beta}$$ $$=\frac{\sin\beta-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin\gamma-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin\alpha-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin(\pi-(\alpha+\gamma))-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin(\pi-(\alpha+\beta))-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin(\pi-(\beta+\gamma))-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin\alpha\cos\gamma+\cos\alpha\sin\gamma-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin \beta\cos\gamma+\cos\beta\sin\gamma-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin\gamma\cos\alpha-\cos\gamma\sin\alpha}{\sin\alpha\sin\gamma}+\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\sin\beta\sin\alpha}+\frac{\sin \beta\cos\gamma-\cos\beta\sin\gamma}{\sin\gamma\sin\beta}$$
$$=(\cot\alpha-\cot\gamma)+(\cot\beta-\cot\alpha)+(\cot\gamma-\cot\beta)$$ $$=0$$

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First lets consider $$\frac{b-2a\cos\gamma}{a\sin\gamma}$$ The numerator looks similar to the RHS of the cosine law, but not quite. It would be nice to see $-2ab\cos\gamma$ instead of $-2a\cos\gamma$. So lets just multiply by $b$. Then the numerator would look like this $$b^2-2ab\cos\gamma$$ Now all that is missing is the $a^2$, so just add it $$a^2+b^2-2ab\cos\gamma$$ Now my expression is equal to $c^2$. Easy, right?

It would be nice if math was really like this, but unfortunately, we can't just add and multiply arbitrary constants whenever it seems convenient to do so. We can, however, add zero (e.g. $+\phi-\phi$) and multiply by one (e.g. $\frac{\phi}{\phi}$).

$$\frac{b-2a\cos\gamma}{a\sin\gamma} = \frac{b-2a\cos\gamma}{a\sin\gamma}\cdot\frac{b}{b}=\frac{b^2-2ab\cos\gamma}{ab\sin\gamma}=\frac{a^2+b^2-2ab\cos\gamma-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}$$ We have just added zero ($a^2-a^2$), so next, lets multiply by one ($\frac{c}{c}$), so we can isolate $\frac{c}{\sin\gamma}$. $$\frac{c^2-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}\cdot\frac{c}{c}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}$$ Keeping in mind, of course that $$\Psi=\frac{c}{\sin\gamma}=\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}$$ Then our expression reduces to $$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\cos\alpha}{b\sin\alpha}+\frac{a-2c\cos\beta}{c\sin\beta}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}+\frac{a^2-b^2}{abc}\cdot\frac{a}{\sin\alpha}+\frac{b^2-c^2}{abc}\cdot\frac{b}{\sin\beta}$$ $$=\frac{\Psi}{abc}\big((c^2-a^2)+(a^2-b^2)+(b^2-c^2)\big)=0$$

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The law of cosines and sines are to be used, your guess is right.

At first consolidate the cosines by summing up with circular symmetry

$$ c^2=a^2+b^2-2ab\cos\gamma + two more $$ gives

$$ a^2 + b^2 + c^2 = 2 ( b\, c \cos\gamma + ... + ....) \tag {1} $$

For sines consolidation it needs multiplying first term numerator and denominator by D where $ c/ \sin \gamma = D, $ the circumcircle diameter.

The first given term becomes

$$ \dfrac{D}{a\,c}(...) $$

All numerators and all denominators can be added separately and placed as numerator and denminator of an equivalent fraction like in:

$$ \frac pq = \frac rs = \frac tu = \frac {p+r+t}{q+s+u} $$

Combine this result with (1) to simplify it to zero.

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Step 1:

Write cosines relations for a triangle:

$$a^2=b^2+c^2-2bc \cos(A)$$ $$b^2=a^2+c^2-2ac \cos(B)$$ $$c^2=a^2+b^2-2ab \cos(C)$$

Step 2:

Write sinus Area relation for the triangle:

$$Area(ABC)=\frac{ab \sin(C)}{2}=\frac{bc \sin(A)}{2}=\frac{ac \sin(B)}{2}$$

Step 3:

Write cosines relations as you have in your question:


$$a^2-b^2=c(c-2b \cos(A))$$ $$c-2b \cos(A)=\frac{a^2-b^2}{c}$$


$$b^2-c^2=a(a-2c \cos(B))$$ $$a-2c \cos(B)=\frac{b^2-c^2}{a}$$


$$c^2-a^2=b(b-2a \cos(C))$$ $$b-2a \cos(C)=\frac{c^2-a^2}{b}$$

Step 4:

Combine with sinus Area relation


$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$ $$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{c b \sin(A) }$$


$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$ $$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{ac \sin(B)}$$


$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$ $$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{ab \sin(C)}$$

Step 5:

Combine with sinus Area relation

$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{2.Area(ABC) }$$ $$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{2.Area(ABC)}$$ $$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{2.Area(ABC)}$$

Step 6:

Add them

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2}{2.Area(ABC) }+\frac{b^2-c^2}{2.Area(ABC)}+\frac{c^2-a^2}{2.Area(ABC)}$$

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{2.Area(ABC) }$$

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=0$$

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