## Calculus: Early Transcendentals 8th Edition

$sin (\alpha-\beta)=sin \alpha cos\beta- cos\alpha sin\beta$
Use the addition formula for cosine and sine identities. $cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$ Change $\alpha$ with $(\pi/2-\alpha)$ $cos [(\pi/2-\alpha)+\beta]= cos (\pi/2-\alpha) cos\beta- sin (\pi/2-\alpha) sin\beta$ $cos [\pi/2-(\alpha-\beta)]= cos (\pi/2-\alpha) cos\beta- sin (\pi/2-\alpha) sin\beta$ Hence, $sin (\alpha-\beta)=sin \alpha cos\beta- cos\alpha sin\beta$