Answer:
[tex]sin(A-B)= -\frac{33}{65}[/tex]
Step-by-step explanation:
step 1
Find the value of sin(A)
we know that
[tex]sin^2(A)+cos^2(A)=1[/tex]
we have
[tex]cos(A)=\frac{3}{5}[/tex]
substitute
[tex]sin^2(A)+(\frac{3}{5})^2=1[/tex]
[tex]sin^2(A)+\frac{9}{25}=1[/tex]
[tex]sin^2(A)=1-\frac{9}{25}[/tex]
[tex]sin^2(A)=\frac{16}{25}[/tex]
[tex]sin(A)=\pm\frac{4}{5}[/tex]
Remember that the angle A is in Quadrant IV
so
The value of sin(A) is negative
therefore
[tex]sin(A)=-\frac{4}{5}[/tex]
step 2
Find the value of sin(B)
we know that
[tex]sin^2(A)+cos^2(A)=1[/tex]
we have
[tex]cos(B)=\frac{12}{13}[/tex]
substitute
[tex]sin^2(B)+(\frac{12}{13})^2=1[/tex]
[tex]sin^2(B)+\frac{144}{169}=1[/tex]
[tex]sin^2(B)=1-\frac{144}{169}[/tex]
[tex]sin^2(B)=\frac{25}{169}[/tex]
[tex]sin(B)=\pm\frac{5}{13}[/tex]
Remember that the angle B is in Quadrant IV
so
The value of sin(B) is negative
therefore
[tex]sin(B)=-\frac{5}{13}[/tex]
step 3
Find the value of sin(A-B)
we know that
[tex]sin(A-B)= sinAcosB-cosAsinB[/tex]
substitute the given values
[tex]sin(A-B)= (-\frac{4}{5})(\frac{12}{13})-(\frac{3}{5})(-\frac{5}{13})[/tex]
[tex]sin(A-B)= (-\frac{48}{65})+(\frac{15}{65})[/tex]
[tex]sin(A-B)= -\frac{33}{65}[/tex]
