Use the given information to evaluate sin (a+b). Exact answers only. No decimals.
Tan a = 4/3, cos b ='-12/13; neither a nor b are in quadrant III.
To evaluate sin (a+b), we need to use one of the trigonometric identities, specifically the sum-of-angles formula for sine:
sin (a+b) = sin a cos b + cos a sin b
Given that tan a = 4/3 and cos b = -12/13, and both a and b are not in quadrant III, we can determine the values of sin a and sin b as follows:
Since tan a = 4/3, we can use the Pythagorean identity to find sin a. The Pythagorean identity states that sin^2 a + cos^2 a = 1. Solving for sin a, we have:
sin^2 a = 1 - cos^2 a
sin^2 a = 1 - (4/5)^2
sin^2 a = 1 - 16/25
sin^2 a = 9/25
sin a = ± √(9/25)
sin a = ± 3/5
Since a is not in quadrant III, the sin a is positive. Therefore, sin a = 3/5.
Similarly, since cos b = -12/13, we can use the Pythagorean identity to find sin b. Solving for sin b, we have:
sin^2 b = 1 - cos^2 b
sin^2 b = 1 - (-12/13)^2
sin^2 b = 1 - 144/169
sin^2 b = 25/169
sin b = ± √(25/169)
sin b = ± 5/13
Since b is not in quadrant III, the sin b is positive. Therefore, sin b = 5/13.
Now, we can substitute the values of sin a, sin b, cos b, and cos a into the sum-of-angles formula for sine:
sin (a+b) = sin a cos b + cos a sin b
sin (a+b) = (3/5)(-12/13) + cos a (5/13)
To evaluate cos a, we can use the Pythagorean identity:
cos^2 a = 1 - sin^2 a
cos^2 a = 1 - (3/5)^2
cos^2 a = 1 - 9/25
cos^2 a = 16/25
cos a = ± √(16/25)
cos a = ± 4/5
Since a is not in quadrant III, the cos a is positive. Therefore, cos a = 4/5.
Substituting the values of cos a, sin a, sin b, and cos b into the expression:
sin (a+b) = (3/5)(-12/13) + (4/5)(5/13)
sin (a+b) = -36/65 + 20/65
sin (a+b) = -16/65
Therefore, sin (a+b) = -16/65.