Question

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

Answer 1

Answer:

Part 1) $sin(A + B) =\frac{185}{697}$

Part 2) $sin(A - B) =\frac{455}{697}$

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

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

we have

$cos(A)=\frac{9}{41}$

substitute

$sin^2(A)+(\frac{9}{41})^2=1$

$sin^2(A)+\frac{81}{1,681}=1$

$sin^2(A)=1-\frac{81}{1,681}$

$sin^2(A)=\frac{1,600}{1,681}$

square root both sides

$sin(A)=\pm\frac{40}{41}$

Angle A is in Quadrant IV

so

sine(A) is negative

therefore

$sin(A)=-\frac{40}{41}$

step 2

Find sin(B)

we know that

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

we have

$cos(B)=-\frac{8}{17}$

substitute

$sin^2(B)+(-\frac{8}{17})^2=1$

$sin^2(B)+\frac{64}{289}=1$

$sin^2(B)=1-\frac{64}{289}$

$sin^2(B)=\frac{225}{289}$

square root both sides

$sin(B)=\pm\frac{15}{17}$

Angle B is in Quadrant III

so

sine(B) is negative

therefore

$sin(B)=-\frac{15}{17}$

step 3

Find sin(A+B)

we know that

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

we have

$sin(A)=-\frac{40}{41}$

$cos(A)=\frac{9}{41}$

$cos(B)=-\frac{8}{17}$

$sin(B)=-\frac{15}{17}$

substitute

$sin(A + B) =(-\frac{40}{41})(-\frac{8}{17}) +(\frac{9}{41})(-\frac{15}{17})$

$sin(A + B) =\frac{320}{697} -\frac{135}{697}$

$sin(A + B) =\frac{185}{697}$

Part 2) Find sin(A-B)

we know that

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

we have

$sin(A)=-\frac{40}{41}$

$cos(A)=\frac{9}{41}$

$cos(B)=-\frac{8}{17}$

$sin(B)=-\frac{15}{17}$

substitute

$sin(A - B) =(-\frac{40}{41})(-\frac{8}{17}) -(\frac{9}{41})(-\frac{15}{17})$

$sin(A - B) =\frac{320}{697} +\frac{135}{697}$

$sin(A - B) =\frac{455}{697}$