Question

cosx+cosy=2 then find cos(x-y)

Answer 1

cos x + cos y = 2
2 cos ((x+y)/2) cos ((x-y)/2) = 2

cos ((x- y)/2) = 1 - cos( (x+y)/2)
cos((x-y)/2) = sin²(x+y)
cos²(x-y) = sin²(x+y) + 1
cos(x-y) = √{sin²(x+y)+1}
cos(x-y) = √2-cos²(x+y)
cos(x-y) = √2-(sinx siny + cos x cosy)
Nah I'm getting tired of it. someone solve it on behalf of me.

Answer 2

Answer:

Step-by-step explanation:

cos x + cos y =2

-1 ≤ cos x ≤1  ----- Eqn1

-1 ≤ cos y ≤ 1 -----Eqn2

so , adding equation 1 &2

     -2 ≤ cos x + cos y ≤ 2

for the maximum value ,

cos x = 1 and cos y = 1

x = y = 0°

as per question , cos (x - y) = cos (0-0)

                                             = cos 0° = 1