
William W. answered 11/11/20
Experienced Tutor and Retired Engineer
sin(122°) = sin(180° - 58°)
cos(32°) = cos(90° - 58°)
cos(122°) = cos(180° - 58°)
sin(32°) = sin(90° - 58°)
sin(122°)cos(32°) - cos(122°)sin(32°)
sin(180° - 58°)cos(90° - 58°) - cos(180° - 58°)sin(90° - 58°)
Using the sin and cos addition and subtraction identities:
sin(x – y) = sin(x)cos(y) – cos(x)sin(y)
cos(x – y) = cos(x)cos(y) + sin(x)sin(y)
The expression becomes:
[sin(180°)cos(58°) – cos(180°)sin(58°)][cos(90°)cos(58°) + sin(90°)sin(58°)] - [cos(180°)cos(58°) + sin(180°)sin(58°)][sin(90°)cos(58°) – cos(90°)sin(58°)] =
[(0)cos(58°) – (-1)sin(58°)][(0)cos(58°) + (1)sin(58°)] - [(-1)cos(58°) +(0)sin(58°)][(1)cos(58°) – (0)sin(58°)] =
sin(58°)sin(58°) - (-cos(58°)cos(58°) =
sin2(58°) + cos2(58°) = 1

William W.
Just realized this is not what was asked for. You should use the sin(x – y) = sin(x)cos(y) – cos(x)sin(y) to re-write the expression as sin(122 - 32) = sin(90) = 111/11/20