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I am looking at the solution of this problem:
$$\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \left(\gamma \right)=4\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)\sin \left(\frac{\gamma }{2}\right)$$ and it says this in the solution as the first step: $$\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \:\left(\gamma \:\right)=2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)-2\cos \left(\frac{\alpha \:+\beta \:}{2}\right)^2+1$$

I understand that the $\cos \alpha +\cos \beta=2\cos \left({\frac {\alpha +\beta }{2}}\right)\cos \left({\frac {\alpha -\beta }{2}}\right)$ is used, however how is the $cos \:\left(\gamma \:\right)$ transformed? I assume I should use the $\alpha +\beta +\gamma =\pi $ to replace $\gamma$ but can't seem to figure it out? Any help is very much appreciated. x

notallwhowanderarelost
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2 Answers2

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We have $$ \cos \gamma = \cos(\pi-(\alpha+\beta)) = -\cos(\alpha+\beta) = -2 \cos^2\frac{\alpha+\beta}{2} + 1$$ Two formulas are used here: $$ \cos (\pi-x) = -\cos x $$ and $$ \cos (2x) = 2 \cos^2 x -1$$ Later you'll also need $$ \cos(\frac{\pi}{2} -x) = \sin x $$

Adam Latosiński
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  • Thank you so much Adam! I did come to the cos(π−x)=−cosx at one point, however was so lost about what to do next, thank you for the other formula especially and for additional help! Not all heroes wear capes. xx – notallwhowanderarelost Feb 06 '21 at 22:56
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You're missing a $+1$:$$\begin{align}\cos\alpha+\cos\beta+\cos\gamma&=2\cos\tfrac{\alpha+\beta}{2}\cos\tfrac{\alpha-\beta}{2}-\cos(\alpha+\beta)\\&=2\cos\tfrac{\alpha+\beta}{2}\left(\cos\tfrac{\alpha-\beta}{2}-\cos\tfrac{\alpha+\beta}{2}\right)+1\\&=2\sin\tfrac{\gamma}{2}2\sin\tfrac{\alpha}{2}\sin\tfrac{\beta}{2}+1.\end{align}$$

J.G.
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