given cos x=2/9, sin y=-/2

Question

find a. sin(x+y)

b. cos (x+y)

Answer 1

The trigonometric function values are cos(x) = 2/9 and sin(y) = - 1/2.

The function cos(x) is positive, so the angle x lies in first quadrant and fourth quadrant.

Using the Pythagorean identity sin² θ + cos² θ = 1, obtain

sin² (x) + (2/9)² = 1

sin² (x) = 1 - 4/81 = 77/81.

sin (x) = ± √(77/81) = ± √(77)/9.

If the angle x lies in first quadrant then sin (x) = √(77)/9.

If the angle x lies in fourth quadrant then sin (x) = - √(77)/9.

 

The function sin(y) is negetive, so the angle y lies in third quadrant and fourth quadrant.

Using the Pythagorean identity sin² θ + cos² θ = 1, obtain

(- 1/2)² + cos² (y) = 1

cos² (y) = 1 - 1/4 = 3/4.

cos(y) = ± √(3/4) = ± √(3)/2.

If the angle y lies in third quadrant then cos(y) = - √(3)/2.

If the angle y lies in fourth quadrant then cos(y) = √(3)/2.

 

Case 1 : The angle x lies first quadrant and y lies in third quadrant.

(a). sin(x + y) = sin(x) cos(y) + cos(x) sin(y).

                        = [ √(77)/9 ][ - √(3)/2 ] + [ 2/9 ][ - 1/2 ]

                        = - √(231)/18 - 1/9

                        = - [√(231) + 2] / 18.

(b). cos(x + y) = cos(x) cos(y) - sin(x) sin(y).

                        = [ 2/9 ][ - √(3)/2 ] - [ √(77)/9 ][ - 1/2 ]

                        = [ - √(3)/9 ] + [ √(77)/18 ]

                        = [- 2√3) + √(77) ] / 18.

 

Case 2 : The angle x lies first quadrant and y lies in fourth quadrant.

(a). sin(x + y) = sin(x) cos(y) + cos(x) sin(y).

                        = [ √(77)/9 ][ √(3)/2 ] + [ 2/9 ][ - 1/2 ]

                        = √(231)/18 - 1/9

                        = [√(231) - 2] / 18.

 

(b). cos(x + y) = cos(x) cos(y) - sin(x) sin(y).

                        = [ 2/9 ][ √(3)/2 ] - [ √(77)/9 ][ - 1/2 ]

                        = [ √(3)/9 ] + [ √(77)/18 ]

                        = [ 2√3) + √(77) ] / 18.