Solve cos(inverse)x - sin(inverse)x = pi/6.
Let cos(in)x = A cos A =x , sin A = sq.rt.(1- x sq)
sin(in)x = B sin B =x , cos B = sq.rt.(1-x sq)
A -B = pi/6
cos A cos B + sin A sin B = cos pi/6 = (sq.rt.3)/2
x sq.rt.( 1-x sq) + sq.rt.(1-x sq) x = (sq.rt.3)/2
2x sq rt.( 1-x sq) = (sq.rt. 3)/2
4xsq(1-x sq) = 3/4
x = + or - 1/2 x + or - (sq rt 3)/2
When someone has time can please check this....coz a text book gives a differfnt answer.
Thanks so much
Let cos(in)x = A cos A =x , sin A = sq.rt.(1- x sq)
sin(in)x = B sin B =x , cos B = sq.rt.(1-x sq)
A -B = pi/6
cos A cos B + sin A sin B = cos pi/6 = (sq.rt.3)/2
x sq.rt.( 1-x sq) + sq.rt.(1-x sq) x = (sq.rt.3)/2
2x sq rt.( 1-x sq) = (sq.rt. 3)/2
4xsq(1-x sq) = 3/4
x = + or - 1/2 x + or - (sq rt 3)/2
When someone has time can please check this....coz a text book gives a differfnt answer.
Thanks so much