Respuesta :

LammettHash
[tex]\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta[/tex]

[tex]\sin\alpha=\dfrac{21}{29}\implies \cos^2\alpha=1-\sin^2\alpha=\dfrac{400}{841}[/tex]

Since [tex]\alpha[/tex] lies in quadrant II, we have [tex]\cos\alpha<0[/tex], so

[tex]\cos\alpha=-\sqrt{\dfrac{400}{841}}=-\dfrac{20}{29}[/tex]

[tex]\cos\beta=\dfrac{15}{17}\implies\sin^2\beta=1-\cos^2\beta=\dfrac{64}{289}[/tex]

[tex]\beta[/tex] lies in quadrant I, so [tex]\sin\beta>0[/tex] and

[tex]\sin\beta=\sqrt{\dfrac{64}{289}}=\dfrac8{17}[/tex]

So

[tex]\sin(\alpha-\beta)=\dfrac{21}{29}\dfrac{15}{17}-\left(-\dfrac{20}{29}\right)\dfrac8{17}=\dfrac{475}{493}[/tex]