To prove if sin(2a+b) = sina , then cos(a+b) = 1.

Since sin(2a+b) = sina,

Sin(2a+b) - sina = 0.....(1)

We know that sin A- sin B = 2 cos{(A+B)/2}sin (A-B)/2}

Therefore from (1) , we get:

2 cos {(2a+b+ a)/2 } sin{(2a+b-a)/2} = 0

2 {cos(a+b)/2 } sin {(a+b)/2} = 0.

Therefore we can equate cos(a+b)/2 = 0.

If cos (a+b)/2 = 0, then cos (a+b) = 1-2{cos[(a+b)/2]]^2 , as cos2A = 1-2(cos(A/2))^2 .

Therefore cos(a+b) = 1- 2{cos(a+b)/2}^2 = 1- 0 = 1.

Therefore cos(a+b) = 1.

Thus If (cos(2a+b) = sina , then cos(a+b) = 1.

We'll expand cosine of the sum:

cos (a+b) = cos a*cos b - sin a*sin b

From enunciation, we know that:

cos a*cos b - sin a*sin b = 1

cos a*cos b = 1 + sin a*sin b (1)

Now, we'll expand the function sin (2a+b):

sin (2a+b) = sin 2a*cos b + sin b*cos 2a

We'll re-write the factor sin 2a:

sin 2a = sin(a+a) = 2sin a*cos a

We'll re-write the factor cos 2a:

cos 2a = cos (a+a) = 1 - 2(sin a)^2

We'll re-write the sum:

sin (2a+b) = 2sin a*cos a*cos b + sin b*[1 - 2(sin a)^2]

We'll substitute the product cos a*cos b by (1):

sin (2a+b) = 2sin a*(1 + sin a*sin b) + sin b*[1 - 2(sin a)^2]

We'll remove the brackets:

sin (2a+b) = 2sin a + 2(sin a)^2*sin b + sin b - 2(sin a)^2*sin b

**sin (2a+b) = 2sin a + sin b**