Assume: $\alpha + \beta + \gamma = \pi$ (Say, angles of a triangle)

Prove: $\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$

There is already a solution on Math-SE, however I want to avoid using the sum-to-product identity because technically the book I go by hasn't covered it yet.

So, is there a way to prove it with identities only as advanced as $\sin\frac{\alpha}{2}$?

Edit: Just giving a hint will probably be adequate (i.e. what identity I should manipulate).

  • $\begingroup$ Have you covered the identity for $\sin(x+y)$ yet? $\endgroup$ – John Molokach Jun 1 '16 at 22:09
  • $\begingroup$ Yes. Otherwise I wouldn't have been able to deduce $sin\frac{\alpha}{2}$. :) $\endgroup$ – Fine Man Jun 1 '16 at 22:10

You may go the other way around: $$ \cos\frac{\gamma}{2}=\cos\frac{\pi-\alpha-\beta}{2}= \sin\frac{\alpha+\beta}{2}= \sin\frac{\alpha}{2}\cos\frac{\beta}{2}+ \cos\frac{\alpha}{2}\sin\frac{\beta}{2} $$ so the right hand side becomes $$ 4\cos\frac{\alpha}{2}\cos\frac{\beta}{2} \sin\frac{\alpha}{2}\cos\frac{\beta}{2}+ 4\cos\frac{\alpha}{2}\cos\frac{\beta}{2} \cos\frac{\alpha}{2}\sin\frac{\beta}{2} $$ Recalling the duplication formula for the sine we get $$ 2\sin\alpha\cos^2\frac{\beta}{2}+2\sin\beta\cos^2\frac{\alpha}{2} $$ and we can recall $$ 2\cos^2\frac{\delta}{2}=1+\cos\delta $$ to get $$ \sin\alpha+\sin\alpha\cos\beta+\sin\beta+\sin\beta\cos\alpha = \sin\alpha+\sin\beta+\sin(\alpha+\beta)= \sin\alpha+\sin\beta+\sin\gamma $$

  • $\begingroup$ Thanks! On my own, I got to the second line, but I didn't recall the duplication formula. $\endgroup$ – Fine Man Jun 4 '16 at 20:11


i) Let $\alpha + \beta + \gamma = \pi \Rightarrow $

$\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$

ii) Let $\alpha + \beta + \gamma > \pi \Rightarrow $

$\sin\alpha + \sin\beta + \sin\gamma > 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$

iii) Let $\alpha + \beta + \gamma < \pi \Rightarrow $

$\sin\alpha + \sin\beta + \sin\gamma < 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$

For example:

If $\alpha, \beta,\gamma \in \left(0,\frac{\pi}{2}\right] $ then $$\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)= 4\sin(\alpha)\sin(\beta)\sin(\gamma) \Leftrightarrow \alpha+\beta+\gamma=\pi$$


1.) $\alpha+\beta+\gamma = \pi$ - equality holds.

Let $\alpha \ge \frac{\pi}{3}$ , then with increasing $\alpha$ decreases the left side and the right side increases $\Rightarrow $

if $\alpha+\beta+\gamma > \pi$ equality is not achieved.

2.) $\alpha+\beta+\gamma \le \pi$

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$R -$ the radius of the circle circumscribed around $\triangle ABC$

$|AD|=|BD|=|CD|=L \ge R$

$$S_{ACD}+S_{CBD}+S_{BAD} \ge S_{ABC} $$

$$S_{ACD}+S_{CBD}+S_{BAD}=\frac{1}{2}L^2(\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma))$$ $$S_{ABC}=\frac{|AB|\cdot|BC|\cdot|CA|}{4R}=\frac{2L^3}{R} \cdot\sin(\alpha)\cdot\sin(\beta)\cdot\sin(\gamma) \ge 2L^2\cdot\sin(\alpha)\cdot\sin(\beta)\cdot\sin(\gamma)$$

Then : $\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)= 4\sin(\alpha)\sin(\beta)\sin(\gamma)$ at $ \alpha, \beta,\gamma \in \left(0,\frac{\pi}{2}\right] $

is performed only when : $\alpha+\beta+\gamma=\pi$

  • $\begingroup$ Couldn't one say this about any math problem? I don't really see anything helpful that will help me solve this problem. Don't worry, no -1, since it's probably my fault that I don't see it being a helpful hint. $\endgroup$ – Fine Man Jun 2 '16 at 2:16
  • $\begingroup$ @SirJony: I added a similar example $\endgroup$ – Roman83 Jun 2 '16 at 6:54

You can easily solve the question.

What you need to do is to plug

$$\beta=\pi-(\alpha +\gamma)$$

Now put it in your equation and you would get

$$ sin(\alpha) + sin(\pi - (\alpha + \gamma)) + sin(\gamma)$$

$$ sin(\alpha) + sin(\alpha + \gamma) + sin(\gamma)$$

Now open $sin(\alpha + \gamma)$ and take $sin(\alpha)$ common from one side and $sin(\gamma)$ for another side and then you would get

$$ sin(\alpha)(1+cos(\gamma)) + sin(\gamma)(1+cos(\alpha))$$

Write $1+cos(\gamma) = 2cos^2(\gamma/2)$ and use the same for $\alpha$

Now write $sin(\alpha)=2sin(\alpha/2)cos(\alpha/2)$ and do the same thing for $sin(\gamma)$ also.

Now you just need to take suitable things common out of the terms and use the sine addition identity. Hope it helps.

I didn't do the whole proof as you only wanted some hint regarding that.


The reason I commented about the addition formula is that you can prove the identity using only that along with the cofunction identities $$\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta,\quad\text{and}\quad\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta.$$ The proof is quite messy with algebra, so I will only outline it and put in the key steps.

1) Use $\frac x2$ twice in the addition formula for sine to get $$\sin x=2\cos\frac x2\sin\frac x2.$$ Then the LHS becomes $$2\left(\sin\frac \alpha 2\cos\frac\alpha 2+\sin\frac \beta 2\cos\frac\beta 2+\sin\frac \gamma 2\cos\frac\gamma 2\right).$$

2) Use the cofunction identity to replace sine half angle to get $$2\left(\cos\left(\frac {\beta+\gamma}2\right)\cos\frac\alpha 2+\cos\left(\frac {\alpha+\gamma}2\right)\cos\frac\beta 2+\cos\left(\frac {\alpha+\beta}2\right)\cos\frac\gamma 2\right).$$

3) Expanding all the addition formulas for cosine, combining like terms and one again using the sine addition formula (in reverse) eventually gives $$2\left(3\cos\frac\alpha 2\cos\frac\beta 2\cos\frac\gamma 2-\sin\frac\gamma 2\sin\left(\frac {\alpha+\beta}2\right)-\cos\frac\gamma 2\sin\frac\alpha 2\sin\frac\beta 2\right).$$

4) Use cofunction identity on the middle term in parentheses to get $$2\left(3\cos\frac\alpha 2\cos\frac\beta 2\cos\frac\gamma 2-\cos\frac\gamma 2\cos\left(\frac {\alpha+\beta}2\right)-\cos\frac\gamma 2\sin\frac\alpha 2\sin\frac\beta 2\right).$$

5) Expanding the cosine addition formula and combining like terms produces the RHS ☺


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