Question 1174989: 1. Given sinB=5/13 in QII and (6,-8) is on the terminal side of a ,find the exact value of sin (a+B).
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
If sin(B) = 5/13, then cos(B) = -12/13. Use the pythagorean theorem to see this. Draw out a right triangle. Note how cosine is negative in quadrant 2.
The distance from (0,0) to (6,-8) is 10 units. Again you'll need the pythagorean theorem to see why this is the case, or you could use the distance formula (which is a rephrasing of the pythagorean theorem).
We have a right triangle with hypotenuse 10, and the two legs are 6 and 8 units each. Let's say we had a third point at (6,0). This is where the 90 degree angle is located. The distance from (0,0) to (6,0) is 6 units to form the horizontal leg, while the vertical leg is 8 units because it spans from (6,0) to (6,-8)
So,
sin(a) = opposite/hypotenuse
sin(a) = -8/10
sin(a) = -4/5
and
cos(a) = adjacent/hypotenuse
cos(a) = 6/10
cos(a) = 3/5
In quadrant 4, sine is negative and cosine is positive
--------------------------------------------------
We have these four facts
sin(a) = -4/5
cos(a) = 3/5
sin(B) = 5/13
cos(B) = -12/13
Now we use a trig identity to find what we're after
sin(a+B) = sin(a)*cos(B) + cos(a)*sin(B)
sin(a+B) = (-4/5)*(-12/13) + (3/5)*(5/13)
sin(a+B) = 48/65 + 15/65
sin(a+B) = (48+15)/65
sin(a+B) = 63/65
Answer: 63/65
|
|
|
