Question

Find the EXACT value of

sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13
where B is in Quadrant IV. Assume all angles are measured from standard position.

Answer 1

`sin(A-B)= -(33)/(65)`

Explanation:

step 1

Find the value of sin(A)

we know that

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

we have

`cos(A)=(3)/(5)`

substitute

`sin^2(A)+((3)/(5))^2=1`

`sin^2(A)+(9)/(25)=1`

`sin^2(A)=1-(9)/(25)`

`sin^2(A)=(16)/(25)`

`sin(A)=\pm(4)/(5)`

Remember that the angle A is in Quadrant IV

so

The value of sin(A) is negative

therefore

`sin(A)=-(4)/(5)`

step 2

Find the value of sin(B)

we know that

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

we have

`cos(B)=(12)/(13)`

substitute

`sin^2(B)+((12)/(13))^2=1`

`sin^2(B)+(144)/(169)=1`

`sin^2(B)=1-(144)/(169)`

`sin^2(B)=(25)/(169)`

`sin(B)=\pm(5)/(13)`

Remember that the angle B is in Quadrant IV

so

The value of sin(B) is negative

therefore

`sin(B)=-(5)/(13)`

step 3

Find the value of sin(A-B)

we know that

`sin(A-B)= sinAcosB-cosAsinB`

substitute the given values

`sin(A-B)= (-(4)/(5))((12)/(13))-((3)/(5))(-(5)/(13))`

`sin(A-B)= (-(48)/(65))+((15)/(65))`

`sin(A-B)= -(33)/(65)`