cos40 + cos60 + cos120 + cos140

Let us rewrite:

cos40 + cos60 + cos(180-60) + cos(140)

But cos(180-60) = - cos60

==> cos40 + cos60 - cos60 + cos(140)

==> cos40 + cos(140)

We know that:

cosa + cosb = 2cos(a+b)/2* cos(a-b)/2

cos 40 + cos(140) = 2cos(90)*cos(-50) = 0

==> cos40 + cos60 + cos120 + cos140 = 0

cos140 = cos(180-40) = - cos4o

cos 120 = cos(180-60) = - cos 60

Adding, we get:

cos140+cos 120 = -cos40-cos60.

cos40+cos60+cos120+cos140 = 0

To find out the value of the sum, we have to use the formula:

cos a + cos b = 2 cos[(a+b)/2]*cos[(a-b)/2]

We'll group the first and the last term together and the middle terms together:

cos 40 + cos 140 = 2 cos[(40+140)/2]*cos[(40-140)/2]

cos 40 + cos 140 = 2 cos[(180)/2]*cos[(-100)/2]

cos 40 + cos 140 = 2 cos 90*cos (-50)

But cos 90 = 0, so:

cos 40 + cos 140 = 0

We'll group the middle terms together:

cos60 + cos120 = 2 cos[(60+120)/2]*cos[(60-120)/2]

cos60 + cos120 = 2 cos 90*cos (-60)

cos60 + cos120 = 0

So, the value of the sum is:

**cos40 + cos60 + cos120 + cos140 = 0+0 = 0**