inverse trigonometry

Asked by  | 13th Jun, 2008, 10:15: AM

Expert Answer:

put cos-1 x/a = α1  and cos-1y/b=α2

so α12=α   and cos α1 =x/a  and cosα2 = y/b

now cos(α12)= 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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