prove that sina=sin(2a+b) if cos(a+b)=1

We'll start by expanding the 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

We'll add sin a*sin b both
sides:

cos a*cos b = sin a*sin b + 1
(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 =
0

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

So, the final result is: sin (2a+b) = 2sin
a + sin b