Answer:
Part 1) [tex]sin(A + B) =\frac{185}{697}[/tex]
Part 2) [tex]sin(A - B) =\frac{455}{697}[/tex]
Step-by-step explanation:
Suppose A is in Quadrant IV and B is in Quadrant III.
If cos(A)=(9/41) and cos(B)=-(8/17)
Part 1) Find sin(A+B)
step 1
Find sin(A)
we know that
[tex]sin^2(A)+cos^2(A)=1[/tex]
we have
[tex]cos(A)=\frac{9}{41}[/tex]
substitute
[tex]sin^2(A)+(\frac{9}{41})^2=1[/tex]
[tex]sin^2(A)+\frac{81}{1,681}=1[/tex]
[tex]sin^2(A)=1-\frac{81}{1,681}[/tex]
[tex]sin^2(A)=\frac{1,600}{1,681}[/tex]
square root both sides
[tex]sin(A)=\pm\frac{40}{41}[/tex]
Angle A is in Quadrant IV
so
sine(A) is negative
therefore
[tex]sin(A)=-\frac{40}{41}[/tex]
step 2
Find sin(B)
we know that
[tex]sin^2(B)+cos^2(B)=1[/tex]
we have
[tex]cos(B)=-\frac{8}{17}[/tex]
substitute
[tex]sin^2(B)+(-\frac{8}{17})^2=1[/tex]
[tex]sin^2(B)+\frac{64}{289}=1[/tex]
[tex]sin^2(B)=1-\frac{64}{289}[/tex]
[tex]sin^2(B)=\frac{225}{289}[/tex]
square root both sides
[tex]sin(B)=\pm\frac{15}{17}[/tex]
Angle B is in Quadrant III
so
sine(B) is negative
therefore
[tex]sin(B)=-\frac{15}{17}[/tex]
step 3
Find sin(A+B)
we know that
[tex]sin(A + B) = sin A cos B + cos A sin B[/tex]
we have
[tex]sin(A)=-\frac{40}{41}[/tex]
[tex]cos(A)=\frac{9}{41}[/tex]
[tex]cos(B)=-\frac{8}{17}[/tex]
[tex]sin(B)=-\frac{15}{17}[/tex]
substitute
[tex]sin(A + B) =(-\frac{40}{41})(-\frac{8}{17}) +(\frac{9}{41})(-\frac{15}{17})[/tex]
[tex]sin(A + B) =\frac{320}{697} -\frac{135}{697}[/tex]
[tex]sin(A + B) =\frac{185}{697}[/tex]
Part 2) Find sin(A-B)
we know that
[tex]sin(A- B) = sin A cos B-cos A sin B[/tex]
we have
[tex]sin(A)=-\frac{40}{41}[/tex]
[tex]cos(A)=\frac{9}{41}[/tex]
[tex]cos(B)=-\frac{8}{17}[/tex]
[tex]sin(B)=-\frac{15}{17}[/tex]
substitute
[tex]sin(A - B) =(-\frac{40}{41})(-\frac{8}{17}) -(\frac{9}{41})(-\frac{15}{17})[/tex]
[tex]sin(A - B) =\frac{320}{697} +\frac{135}{697}[/tex]
[tex]sin(A - B) =\frac{455}{697}[/tex]
