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

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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**