alleryStart learning →Given sin a = 3/5 with a in quadrant I and cos b = -5/13 with b in quadrant III, find sin(a + b).
A16 / 65
B63 / 65
C-63 / 65
D-33 / 65
Answer & Solution
Correct answer: C. -63 / 65
1. Build a reference triangle for a with opposite side 3 and hypotenuse 5, so the adjacent side is 4 and cos a = 4/5.
2. Build a reference triangle for b with adjacent side -5 and hypotenuse 13, so the remaining side squares to 169 - 25 = 144.
3. That side is 12 in length, and it is negative in quadrant III, so sin b = -12/13.
4. Apply sin(a + b) = sin a cos b + cos a sin b.
5. The first product is (3/5)(-5/13) = -15/65.
6. The second product is (4/5)(-12/13) = -48/65.
7. Adding gives -63/65.
8. The value 16/65 is cos(a + b) for the same pair of angles, so it answers a different question.
_Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 9 "Trigonometric Identities and Equations", section 9.2 Sum and Difference Identities_
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