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I'm stuck with this from a few hours. There is an exercise in my textbook, which is solved and it's must be used as an example, however I can't understand it. Here's the exercises + how it's solved. Given that: $$ \alpha + \beta + \gamma = \pi $$ proof that $$ \sin(\alpha) + \sin(\beta) + \sin(\gamma) = 4 * \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\gamma}{2}\right) $$

Solution from my textbook: $$ \sin(\alpha) + \sin(\beta) + \sin(\gamma) = 2*\sin\left(\frac{\alpha+ \beta}{2}\right)*\cos\left(\frac{\alpha-\beta}{2}\right) + \sin(\pi - (\alpha + \beta)) $$ $$ = 2*\sin\left(\frac{\alpha+ \beta}{2}\right)*\cos\left(\frac{\alpha-\beta}{2}\right) + 2*\sin\left(\frac{\alpha+\beta}{2}\right)*\cos\left(\frac{\alpha+\beta}{2}\right) $$ $$ = 2*\sin\left(\frac{\alpha+ \beta}{2}\right)*\left(\cos\left(\frac{\alpha-\beta}{2}\right) + \cos\left(\frac{\alpha+\beta}{2}\right)\right) $$ $$ = 4 * \sin\left(\frac{\pi -\gamma}{2}\right) * \cos\left(\frac{\alpha}{2}\right) * cos\left({\frac{\beta}{2}}\right) $$ $$ = 4 * \cos \left(\frac{\alpha}{2}\right) * \cos\left(\frac{\beta}{2}\right) * \cos\left(\frac{\gamma}{2}\right) $$

The thing I can't understand is how from $$ \sin(\pi - (\alpha + \beta)) $$

Become

$$ 2*\sin\left(\frac{\alpha+\beta}{2}\right)*\cos\left(\frac{\alpha+\beta}{2}\right) $$

I think this is the formula for $$ \sin(2*\alpha) $$

mathlove
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user245878
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  • $\sin(\pi - (\alpha + \beta)) =2\sin(\frac{\pi}{2}-\frac{\alpha+\beta}{2}) \cos(\frac{\pi}{2}-\frac{\alpha+\beta}{2}) =2\cos(\frac{\alpha+\beta}{2}) \sin(\frac{\alpha+\beta}{2})$ – Alexey Burdin Jun 04 '15 at 18:51

2 Answers2

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Using$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$$$\sin(2C)=2\sin C\cos C$$gives you$$\begin{align}\sin\left(\pi-(\alpha+\beta)\right)&=\sin\pi\cos(\alpha+\beta)-\cos\pi\sin(\alpha+\beta)\\&=0\cdot \cos(\alpha+\beta)-(-1)\cdot\sin(\alpha+\beta)\\&=\sin(\alpha+\beta)\\&=\sin\left(2\cdot\left(\frac{\alpha+\beta}{2}\right)\right)\\&=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)\end{align}$$

mathlove
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One more observation and you're there.

Note that $\sin(\pi - (\alpha+\beta)) = \sin(\alpha + \beta)$.

Then application of the double-angle formula is clear.

John
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  • I know that $$ \sin(\pi - (\alpha + \beta) = \sin(\alpha + \beta)$$ However it's never said taht $$ \alpha = \beta$$ – user245878 Jun 04 '15 at 18:56
  • Why do you need $\alpha=\beta$? The part you said you were not understanding doesn't depend on $\alpha$ equaling $\beta$. – John Jun 04 '15 at 18:59
  • I don't understand how the formula for double-angle is applied here since it's about $$ 2 *\alpha$$ – user245878 Jun 04 '15 at 19:00