# Calculate the sum sin120+sin170+sin190+sin240

hala718 | Certified Educator

sin 120 + sin 170 + sin 190 + sin 240 =

Let us rewrite:

sin (180-60) + sin (180 - 10) + sin (180+10) + sin (180 + 60)

sin(180 - 60) = sin 60  in 2nd quadrant  ( positive)

sin (180 - 10) = sin 10  (positive in 2nd quadrant)

sin (180 + 10) = - sin 10  ( 3rd quadrant negative)

sin (180 + 60) = - sin 60 (3rd quadrant negative)

Now ley us substitutue:

sin 60 + sin 10  - sin 10 - sin 60 = 0

giorgiana1976 | Student

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

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

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

sin 120 + sin 240 = 2 sin[(120+240)/2]*cos[(120-240)/2]

sin 120 + sin 240 = 2 sin[(360)/2]*cos[(-120)/2]

sin 120 + sin 240= 2 sin 180*cos (-60)

But sin 180 = 0, so:

sin 120 + sin 240 = 0

We'll group the middle terms together:

sin170+sin190 = 2 sin[(170+190)/2]*cos[(170-190)/2]

sin170+sin190 = 2 sin180*cos (-10)

sin170+sin190 = 0

So, the value of the sum is:

sin120+sin170+sin190+sin240 = 0+0 = 0

neela | Student

Re arranging the given expression we get :

(sin120+sin240)+(in170+sin190)..........(1)

We know that:

sin120 = sin(180-60) = sin60

sin240 =sin(180+60) = -sin60

sin179 = sin(180-10) = sin10

sin(190) = sin(180+10) = sin 10

Therefor by substitution in  (1) we get:

(sin60-sin60)+(sin10 - sin10) = 0