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lions [1.4K]
3 years ago
12

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

Mathematics
1 answer:
Arturiano [62]3 years ago
5 0

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}

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