cosx+cosy=2 then find cos(x-y)
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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.
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.
Answered by
1
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
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