## Calculus: Early Transcendentals 8th Edition

$cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$
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, $cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$