inverse trigonometry
Asked by
| 13th Jun, 2008,
10:15: AM
Expert Answer:
put cos-1 x/a = α1 and cos-1y/b=α2
so α1+α2=α and cos α1 =x/a and cosα2 = y/b
now cos(α1+α2)= cosα1*cosα2 - sinα1*sinα2 = cosα
or cosα1*cosα2-cosα=sinα1*sinα2
square both the sides and replace the values
sinα1 = ( 1- cos2α1)1/2 = (1- (x2/a2))1/2
sinα2 = ( 1- cos2α2)1/2 = (1- (y2/b2))1/2
Answered by
| 10th Aug, 2008,
03:53: PM
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