Answer
$ cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$
Work Step by Step
Apply the distance formula for $(cos \alpha, sin\alpha),(cos \beta, sin\beta)$
$c^{2}=(cos \beta-cos \alpha)^{2}+(sin \beta-sin \alpha)^{2}$
$c^{2}=(cos \beta-cos \alpha)^{2}+(sin \beta-sin \alpha)^{2}$
$c^{2}=2-2 cos \alpha cos\beta-2 sin \alpha sin\beta$
Using the law of cosines, we have
$c^{2}=a^{2}+b^{2}-2ab cos \theta$
Here, $c=c, a=1, b=1, \theta=\alpha-\beta$
Thus, $c^{2}=2-2 cos (\alpha-\beta)$ ...(2)
From equations (1) and (2), we have
$2-2 cos (\alpha-\beta)=2-2 cos \alpha cos\beta-2 sin \alpha sin\beta$
$2 cos (\alpha-\beta)=2 cos \alpha cos\beta+2 sin \alpha sin\beta$
$ cos (\alpha-\beta)= cos \alpha cos\beta+ sin \alpha sin\beta$
Change the sign of $\beta$ with $-\beta$.
$ cos (\alpha+\beta)= cos \alpha cos(-\beta)+ sin \alpha sin(-\beta)$
Hence, $ cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$