Here is a problem from Gelfand's Trigonometry:

Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta-\gamma}{2}\right).$$

I have tried to worked through this problem but cannot complete it. If I let $A= \alpha -\beta$, $B=\beta-\gamma$ and $C= \beta-\gamma$, and $A+B+C=\pi$ (now $A$, $B$ and $C$ are angles of a triangle), then I could prove the equality. But without this condition, I am stuck.

Could you show me how to complete this exercise?

  • 1
    $\begingroup$ Hint: Use Half-angle formulae $\endgroup$
    – tien lee
    May 11, 2018 at 0:47
  • 2
    $\begingroup$ I think you have a typo; your $B$ and $C$ are defined the same way. $\endgroup$
    – BallBoy
    May 11, 2018 at 1:01
  • $\begingroup$ Yeah, sorry, definitely typo. $\endgroup$ May 11, 2018 at 1:04

3 Answers 3


$$ \begin{align} \color{#C00}{\sin(x)+\sin(y)}+\color{#090}{\sin(x+y)} &=\color{#C00}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)}+\color{#090}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x+y}2\right)}\\ &=2\sin\left(\frac{x+y}2\right)\left[\cos\left(\frac{x-y}2\right)+\cos\left(\frac{x+y}2\right)\right]\\ %&=2\sin\left(\frac{x+y}2\right)\,\color{#00F}{2\cos\left(\frac x2\right)\cos\left(\frac y2\right)}\\ %&=4\sin\left(\frac{x+y}2\right)\cos\left(\frac x2\right)\cos\left(\frac y2\right) \end{align} $$ Finish off by using the formula for the cosine of a sum/difference, then set $x=\alpha-\beta$ and $y=\beta-\gamma$.

  • $\begingroup$ Much neater than my solution. $\ddot \smile$ $\endgroup$ May 11, 2018 at 1:32
  • $\begingroup$ Thank you very much, your derivation is beautiful. I have known your name before, Robjohn. $\endgroup$ May 11, 2018 at 1:44

Use \begin{eqnarray*} \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \end{eqnarray*} to give \begin{eqnarray*} \sin (\alpha-\beta) + \sin (\alpha-\gamma) = 2 \sin \left( \frac{2 \alpha-\beta-\gamma}{2} \right) \cos \left( \frac{\beta-\gamma}{2} \right). \end{eqnarray*} Now use the double angle formula \begin{eqnarray*} \sin(\beta-\gamma)=2 \sin \left(\frac{\beta-\gamma}{2} \right) \cos \left(\frac{\beta-\gamma}{2}\right). \end{eqnarray*} Use the first formula again & the result follows.

  • $\begingroup$ I have used that formula for all possible combinations of three terms already. The result is not so. For example, following your suggestion, I will have: s$2sin(\frac{2\alpha-\beta-\gamma}{2})cos(\frac{\gamma-\beta}{2})+2sin(\frac{\beta-\gamma}{2})cos(\frac{\beta-\gamma}{2})$. This does not lead to the desired result. $\endgroup$ May 11, 2018 at 1:21
  • $\begingroup$ \begin{eqnarray*} \cdots=2 \cos(\frac{\beta-\gamma}{2}) (\sin(\frac{\beta-\gamma}{2}) +\sin \left( \frac{2\alpha-\beta-\gamma}{2} \right))= \cdots \end{eqnarray*} & use the formula $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$. $\endgroup$ May 11, 2018 at 1:29

Use factoring and algebra. Let $\;X:=e^{ix},\;Y:=e^{iy},\;Z:=e^{iz}.\;$ Then the following equations hold

$$\;\sin(x) = \frac{X-X^{-1}}{2i}, \; \cos(x)=\frac{X+X^{-1}}2,\; \sin(x-y) = \frac{X^2-Y^2}{2iXY},\; \cos(x-y)=\frac{X^2+Y^2}{2XY}.$$

Summing and factoring these equations using a Computer Algebra System gives the equation $$ \sin(x\!-\!y) \!+\! \sin(x\!-\!z) \!+\! \sin(y\!-\!z) \!=\! \frac{X^2\!-\!Y^2}{2iXY} \!+\! \frac{X^2\!-\!Z^2}{2iXZ} \!+\! \frac{Y^2\!-\!Z^2}{2iYZ} \!=\! \frac{(X\!+\!Y)(X\!-\!Z)(Y\!+\!Z)}{2iXYZ}. $$

$$\textrm{Also now we have} \quad\cos\frac{x-y}2 = \frac{X+Y}{2\sqrt{XY}},\quad \sin\frac{x-z}2 = \frac{X-Z}{2i\sqrt{XZ}},\quad \cos\frac{y-z}2 = \frac{Y+Z}{2\sqrt{YZ}},$$ and the result follows. Of course, trigonometric identities can and have also been used to prove it.

With a slight sign change this more symmetrical trigonometric identity is also true

$$ \sin(x\!-\!y) \!+\! \sin(y\!-\!z) \!+\! \sin(z\!-\!x) \!=\! -4 \sin\frac{x-y}2 \sin\frac{y-z}2 \sin\frac{z-x}2 . $$


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