a sin(x) + b cos( x + y ) + b cos( x - y ) = d

or, a sin(x) + b [ 2 cos(x) cos(y) ] = d

or, cos(y) = [ d - a sin(x) ] / 2 b cos(x)

or, cos(y) = [ d - a sin(x) ] / 2 b (1 - sin²x)^1/2

Now first term is minimum when sin(x) = 0 and second term is minimum when sin(x) = 1

So, min (cos(y)) = (d / 2b) - 0 = d / 2b

min( |cos(y)| = | d / 2b |

in last image that you have attached, second term is wrong i think. please check. and how can you pur different values of sinx together.

yeah thnx himsy..i was in a hurry nd didn't noticed this one.

I am continuing the question from second last step now,

cos(y) = ( sec(x) ) [ ( d / 2b ) - ( a / 2b ) ( sin(x) ) ]

This value will be minimum when sin(x) is maximum i.e. 1 but when sinx = 1, secx is not defined.

So, this expression has no global minima according to me.

Nudge me if you still have any problem..........

how did you get (a/2b)sinx? tell me how did you upload the picture? i will tell the answer.