may i have some help?

Question

find sin (a-b) if sin (a)= 3/5 where a is in the second quadrant  and cos (b)= 15/17 where b is in the first quadrant

 

1.) 36/85

 

2.) 84/85

 

3.) 13/85

 

4.) 77/85

Answer 1

The difference formula:  sin(a-b) = sin(a) cos(b) - cos(a) sin(b)  is helpful here.

To use it we will need cos(a) and sin(b).

The triangle associated with angle a is the 3,4,5 right triangle.  From this we can see that cos(a)

will be either 4/5 or -4/5, but since we are told that a is in the second quadrant, we have  cos(a) = -4/5.

The triangle associated with angle b is the 8,15,17 right triangle.   Since b is in the first quadrant,

we see right away that sin(b) = 8/17.

Thus sin(a-b) = (3/5)(15/17) - (-4/5) (8/17) =   77/(5 x 17)

By looking at cos(a-b), it could be figured out that a-b is in the second quadrant.

 

 

Answer 2

sin(a-b) can be rewritten as the following

 

sin(a-b) = sina cosb - cos a sinb

 

we know sin a = 3/5     and a is in the second quadrant

 

therefore cos a = -4/5   (draw a triangle in the second quadrant with one vertex at (0,0) the second at (-4,0) and the third at (-4,3)

 

 

cos b = 15/17 and b is in the first quadrant  

 

therefore 

 

sin b = 8/17  (figure this out by drawing triangle in the first quadrant)

 

and now we have

 

sin (a-b) = 3/5 * 15/17 - (-4/5) * 8/17

 

            = 45/85 - (-32/85)

 

            = 77/85

Answer 3

sin(a-b)=sin(a)cos(b)-cos(a)sin(b)

 

so we need to find cos(a) and sin(b) first. Knowing a is in the second quadrant, cos(a) would be negative and b in the first quadrant, sin(b) would be positive.

 

cos(a) = -√(1-sin(a)²) = -√(1-(3/5)²) = -4/5

sin(b)= √(1-cos(b)²) = √(1-(15/17)²) = 8/17

 

using all the information and plugging it in the first equation, we'll have:

 

sin(a-b) = (3/5)(15/17)-(-4/5)(8/17)

sin(a-b) = 45/85 + 32/85

sin(a-b) = 77/85

 

which is the answer #4