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Question Number 126720 by mathocean1 last updated on 23/Dec/20

$$$$$$showthat$$$$cos\left(a\right)+cos\left(b\right)=2cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)$$

Answered by mathmax by abdo last updated on 23/Dec/20

$$\mathrm{let}\frac{\mathrm{a}+\mathrm{b}}{2}=\mathrm{u}\mathrm{and}\frac{\mathrm{a}-\mathrm{b}}{2}=\mathrm{v}\Rightarrow \{\begin{array}{l}\mathrm{a}+\mathrm{b}=2\mathrm{u}\\ \mathrm{a}-\mathrm{b}=2\mathrm{v}\Rightarrow \end{array}$$$$\{\begin{array}{l}\mathrm{a}=\mathrm{u}+\mathrm{v}\\ \mathrm{b}=\mathrm{u}-\mathrm{v}\Rightarrow \mathrm{coaa}+\mathrm{cosb}=\cos (\mathrm{u}+\mathrm{v})+\cos (\mathrm{u}-\mathrm{v})\end{array}$$$$=\mathrm{cosu}\mathrm{cosv}-\mathrm{sinu}\mathrm{sinv}+\mathrm{cosu}\mathrm{cosv}+\mathrm{sinu}\mathrm{sinv}$$$$=2\mathrm{cosu}\mathrm{cosv}=2\cos \left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\cos \left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)\mathrm{the}\mathrm{equality}\mathrm{is}\mathrm{proved}$$

Answered by mr W last updated on 24/Dec/20

$$\theta =\frac{\theta +\phi }{2}+\frac{\theta -\phi }{2}$$$$\phi =\frac{\theta +\phi }{2}-\frac{\theta -\phi }{2}$$$$\cos \theta =\cos (\frac{\theta +\phi }{2}+\frac{\theta -\phi }{2})=\cos \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}-\sin \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2}$$$$\cos \phi =\cos (\frac{\theta +\phi }{2}-\frac{\theta -\phi }{2})=\cos \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}+\sin \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2}$$$$\Rightarrow \cos \theta +\cos \phi =2\cos \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}$$

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