Question

Suppose A is inQuadrant IV and B is in Quadrant III.

If cos(A)=(9/41) and cos(B)=-(8/17), then sin(A+B)=
Compute sin (A-B)=

Answer 1

Part 1)
`sin(A + B) =(185)/(697)`

Part 2)
`sin(A - B) =(455)/(697)`

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

`sin^2(A)+cos^2(A)=1`

we have

`cos(A)=(9)/(41)`

substitute

`sin^2(A)+((9)/(41))^2=1`

`sin^2(A)+(81)/(1,681)=1`

`sin^2(A)=1-(81)/(1,681)`

`sin^2(A)=(1,600)/(1,681)`

square root both sides

`sin(A)=\pm(40)/(41)`

Angle A is in Quadrant IV

so

sine(A) is negative

therefore

`sin(A)=-(40)/(41)`

step 2

Find sin(B)

we know that

`sin^2(B)+cos^2(B)=1`

we have

`cos(B)=-(8)/(17)`

substitute

`sin^2(B)+(-(8)/(17))^2=1`

`sin^2(B)+(64)/(289)=1`

`sin^2(B)=1-(64)/(289)`

`sin^2(B)=(225)/(289)`

square root both sides

`sin(B)=\pm(15)/(17)`

Angle B is in Quadrant III

so

sine(B) is negative

therefore

`sin(B)=-(15)/(17)`

step 3

Find sin(A+B)

we know that

`sin(A + B) = sin A cos B + cos A sin B`

we have

`sin(A)=-(40)/(41)`

`cos(A)=(9)/(41)`

`cos(B)=-(8)/(17)`

`sin(B)=-(15)/(17)`

substitute

`sin(A + B) =(-(40)/(41))(-(8)/(17)) +((9)/(41))(-(15)/(17))`

`sin(A + B) =(320)/(697) -(135)/(697)`

`sin(A + B) =(185)/(697)`

Part 2) Find sin(A-B)

we know that

`sin(A- B) = sin A cos B-cos A sin B`

we have

`sin(A)=-(40)/(41)`

`cos(A)=(9)/(41)`

`cos(B)=-(8)/(17)`

`sin(B)=-(15)/(17)`

substitute

`sin(A - B) =(-(40)/(41))(-(8)/(17)) -((9)/(41))(-(15)/(17))`

`sin(A - B) =(320)/(697) +(135)/(697)`

`sin(A - B) =(455)/(697)`