25.5k views
0 votes
Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2

1 Answer

3 votes

Answer:

The work is in the explanation.

Explanation:

The sine addition identity is:


\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).

The sine difference identity is:


\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a).

The cosine addition identity is:


\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b).

The cosine difference identity is:


\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b).

We need to find a way to put some or all of these together to get:


\sin(a)\cos(b)=(\sin(a+b)+\sin(a-b))/(2).

So I do notice on the right hand side the
\sin(a+b) and the
\sin(a-b).

Let's start there then.

There is a plus sign in between them so let's add those together:


\sin(a+b)+\sin(a-b)


=[\sin(a+b)]+[\sin(a-b)]


=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]

There are two pairs of like terms. I will gather them together so you can see it more clearly:


=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]


=2\sin(a)\cos(b)+0


=2\sin(a)\cos(b)

So this implies:


\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)

Divide both sides by 2:


(\sin(a+b)+\sin(a-b))/(2)=\sin(a)\cos(b)

By the symmetric property we can write:


\sin(a)\cos(b)=(\sin(a+b)+\sin(a-b))/(2)

User DotNetStudent
by
8.6k points