A,B are 2 different values of x between 0&180 which satisfies 6cosx+8sin=9.Find sin(a+B).

Asked by  | 4th Oct, 2008, 12:48: PM

Expert Answer:

divide the eqn by (62 + 82) = 10 both side we get 6/10(cosx)+8/10(sinx)=9/10   let sin m =6/10..==>cos m = 8/10

so eq. becomes sin m cos x  + cos m sin x = 9/10====>  sin(m + x) = 9/10..

this eqn has two soln..===>  x = arc sin 9/10 - m   and  x =  180 - (arc sin 9/10 - m ) so the soln is {where arc is inverse function}

x = A  = arc sin 9/10 - arc sin 6/10     and    x  = B = 180 - (arc sin 9/10 - arc sin 6/10)  

so A+B = 180..=====> sin (A+B)  = sin(180) = 0

Answered by  | 16th Dec, 2008, 09:35: PM

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