Proof of additive property for sine

[jtblizard]

Homework Statement

We are supposed to prove that sin(x+y) = cos(x)sin(y) + sin(x)cos(y)

Homework Equations

cos(A-pi/2) - sin(A)
sin(pi/2 - A) = Cos(A)
sin(A-pi/2) = -cos(A)

The Attempt at a Solution

We had to prove all of the relevant equations but were allowed to work in groups and now that I am alone, I am just having trouble pushing off and getting started. If someone can give me a push in the right direction, I feel like I will be able to finish it on my own. Thank you in advance.

 

[Hurkyl]

The cited relevant equations aren't enough to do this proof. Is there anything else useful you can use?

You said you've proven similar things in a group -- what sorts of things? And how were they proven? Can you do something similar?

 

[jtblizard]

We also have the difference property for Cosine: cos(A-B)=cos(A)cos(B)+sin(A)sin(B) and we have that same thing written with x instead of A and once with x instead of A and with y-pi/2 instead of B. The things listed under relevant equations are the things we proved in class. Once, we used a triangle with one angle equal to A in order to prove sin(pi/2 - A) = cos(A) and we know that sin(-x) = -sin(x)

 

[HallsofIvy]

If you know that cos(A- B)= cos(A)cos(B)+ Sin(A)sin(B) then you also know that cos(A+ B)= cos(A- (-B))= cos(A)cos(-B)+ sin(A)sin(-B). And since cos(-B)= cos(B) and sin(-B)= -sin(B), cos(A+ B)= cos(A)cos(B)- sin(A)sin(B). Now use the fact that sin^2(\theta)= 1- cos^2(\theta) with \theta= x+ y.

 

[LCKurtz]

Another approach is to write

\sin{(a+b)} = \cos{(\frac \pi 2 - (a + b))} = \cos{((\frac \pi 2 - a)-b))}

and use the cosine formula.