Question

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

Answer 1

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)`