eliminate x from the equations a cos x+b sin x + c = 0 and a1 cos x+b1 sinx +c1=0 where a,b,c,a1,b1,c1 are real constants and ab1-a1b is not equal to 0
Answers
Answered by
7
Given two equations : asinx + bcosx + c = 0
a1sinx + b1cosx + c1 = 0
Let sinx = X and cosx = Y
then, we can write equations as
aX + bY + c = 0 ........(i)
a1X + b1Y + c1 = 0 .........(ii)
now, solve equations. (i) and (ii) with help of Cramer rule ,
X/(bc1 - b1c) = -Y/(ac1 - a1c) = 1/(ab1 - a1b)
so, X = (bc1 - b1c)/(ab1 -a1b)
and Y = - (ac1 - a1c)/(ab1 - a1b)
put X = sinx and Y = cosx
then, sinx = (bc1 - b1c)/(ab1 -a1b) ....(iii)
cosx = -(ac1 - a1c)/(ab1 - a1b) ......(iv)
dividing equation (iii) by (iv),
sinx/cosx = {(bc1 - b1c)/(ab1 -a1b)}/{-(ac1 - a1c)/(ab1 - a1b)}
tanx = (bc1 - b1c)/(a1c - ac1)
x = arctan{(bc1 - b1c)/(a1c - ac1)}
hence,
a1sinx + b1cosx + c1 = 0
Let sinx = X and cosx = Y
then, we can write equations as
aX + bY + c = 0 ........(i)
a1X + b1Y + c1 = 0 .........(ii)
now, solve equations. (i) and (ii) with help of Cramer rule ,
X/(bc1 - b1c) = -Y/(ac1 - a1c) = 1/(ab1 - a1b)
so, X = (bc1 - b1c)/(ab1 -a1b)
and Y = - (ac1 - a1c)/(ab1 - a1b)
put X = sinx and Y = cosx
then, sinx = (bc1 - b1c)/(ab1 -a1b) ....(iii)
cosx = -(ac1 - a1c)/(ab1 - a1b) ......(iv)
dividing equation (iii) by (iv),
sinx/cosx = {(bc1 - b1c)/(ab1 -a1b)}/{-(ac1 - a1c)/(ab1 - a1b)}
tanx = (bc1 - b1c)/(a1c - ac1)
x = arctan{(bc1 - b1c)/(a1c - ac1)}
hence,
Similar questions