Question

cos40+cos50+cos70+cos80=cos20+cos 10

Answer 1

Given, cos 40 + cos 50 + cos 70 + cos 80

We know that cos a + cos b= 2 cos(a+b)/2 cos(a-b)/2.

        (2 cos(40+80)/2 cos(40-80)/2) + (2 cos (50+70)/2 cos(50-70)/2)

      = 2 cos 60 cos (-20)  + 2 cos 60 cos (-10)

     = 2 * 1/2 * cos 20 + 2 * 1/2 cos 10

     = cos 20 + cos 10.

Answer 2

hello users .......

we have to show that :-
Cos40+cos50+cos70+cos80 = cos20+cos 10

solution :-

We know that: cos c + cos d= 2 cos(c+d)/2 cos(a-d)/2.
here,
(cos 40 + cos 50) + (cos 70 + cos 80) =

{2 cos(40+80)/2 cos(40-80)/2} + {2 cos (50+70)/2 cos(50-70)/2} 

= 2 cos 120 / 2 cos (-40/2)  + 2 cos 120/2 cos (-20 / 2)

=2 cos 60 cos (-20)  + 2 cos 60 cos (-10)

= 2 * 1/2 * cos 20 + 2 * 1/2 cos 10

= cos 20 + cos 10.

hence LHS = RHS