Find the exact value of sin(u+v) given that sin u = 7/25 and cos v= -12/13. Both u and v are in Quadrant II.

Question

Find the exact value of sin(u+v) given that sin u = 7/25 and cos v= -12/13.  Both u and v are in Quadrant II.

Answer 1

The first results are from the Pythagorean identity cos²x + sin²x=1.  cosx=±√(1-sin²x), and sin x =±√(1-cos²x)

For u in the second quadrant, sin u=7/25, cos u <0 and equal to -√(1-(7/25)²)=-24/25.  Or you can just remember the 7, 24, 25 right triangle.

For v in the second quadrant sin v >0 and equal to 5/13, if you remember the 5, 12, 13 right triangle or, as above sin v = √(1-(12/13)²) = 5/13

NOW: sin(u+v)=sin u cos v + cos u sin v =(7/25)(-12/13)+(-24/25)(5/13)=-204/325