# Thread: Product/sum identities - I need enlightenment!

1. ## Product/sum identities - I need enlightenment!

2. Hello, Kaitosan!

I saw these derived many year ago and the approach is still with me.

We use these Compound Angle Identities:

. . $\displaystyle \sin(A \pm B) \:=\:\sin A\cos B \pm \sin B\cos A$

. . $\displaystyle \cos(A \pm B) \:=\:\cos A\cos B \mp \sin A\sin B$

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Product-to-Sum Identities

. . $\displaystyle \begin{array}{cccc} [1] & \cos x\cos y &=& \frac{1}{2}\bigg[\cos(x+y) + \cos(x-y)\bigg] \\ \\[-3mm] [2] & \sin x\sin y &=&\frac{1}{2}\bigg[\cos(x-y) - \cos(x+y)\bigg] \\ \\ [-3mm] [3] & \sin x\cos y &=&\frac{1}{2}\bigg[\sin(x+y) + \sin(x-y)\bigg] \end{array}$

We have: .$\displaystyle \begin{array}{ccccc}\cos(x+y) \:=\:\cos x\cos y - \sin x\sin y & {\color{blue}(a)} \\ \cos(x-y) \:=\:\cos x\cos y + \sin x\sin y & {\color{blue}(b)} \end{array}$

Derivation of [1]

Add $\displaystyle {\color{blue}(a) + (b)}\!:\;\;\cos(x+y) + \cos(x-y) \;=\;2\cos x\cos y$

And we have: .$\displaystyle \cos x\cos y \:=\:\tfrac{1}{2}\bigg[\cos(x+y) + \cos(x-y)\bigg]$

Derivation of [2]

Subtract $\displaystyle {\color{blue}(b) - (a)}\!: \;\;\cos(x-y) - \cos(x+y) \:=\:2\sin x\sin y$

And we have: .$\displaystyle \sin x\sin y \:=\:\tfrac{1}{2}\bigg[\cos(x-y) - \cos(x+y)\bigg]$

Derivation of [3]

We have: .$\displaystyle \begin{array}{cccc} \sin(x+y) \:=\:\sin x\cos y + \sin y\cos x & {\color{blue}(c)} \\ \sin(x-y) \:=\:\sin x\cos y - \sin y\cos x & {\color{blue}(d)} \end{array}$

Add $\displaystyle {\color{blue}(c) + (d)}\!:\;\;\sin(x+y) + \sin(x-y) \:=\:2\sin x\cos y$

And we have: .$\displaystyle \sin x\cos y \:=\:\tfrac{1}{2}\bigg[\sin(x+y) + \sin(x-y)\bigg]$

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Sum-to-Product Identities

. . $\displaystyle \begin{array}{cccc} \sin x + \sin y &=& 2\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \\ \\[-4mm] \sin x - \sin y &=& 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) \\ \\[-4mm] \cos x + \cos y &=& 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \\[-4mm] \cos x - \cos y &=& \text{-}2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) \end{array}$

These can be derived from the previous four identities with a substitution.

Let: .$\displaystyle \begin{array}{ccc} u &=& x+y \\ v &=& x-y \end{array}\;\;{\color{red}[1]}\quad\Rightarrow\quad \begin{array}{ccc}x &=&\dfrac{u+v}{2} \\ \\[-4mm] y &=& \dfrac{u-v}{2} \end{array}\;\;{\color{red}[2]}$

From [3] we have: .$\displaystyle \sin(x+y) + \sin(x-y) \:=\:2\sin x\cos y$

Substitute [1] and [2]: .$\displaystyle \sin u + \sin v \:=\: 2\sin\left(\frac{u+v}{2}\right) \cos\left(\frac{u-v}{2}\right)$

. . . and so on.

3. I see. Thank you.

Personally, I think there's a clearer way to solve the sum identities.

For example...

sinx + siny = ???

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

Substituted x and y for double angles while keeping in mind that for all the sum identities that one trig identity must have a subtraction of angles (in this case, y).

sina cosb + sinb cosa + sina cosb - sinb cosa added