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Thread: Product/sum identities - I need enlightenment!

  1. #1

  2. #2
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    Hello, Kaitosan!

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

    We use these Compound Angle Identities:

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

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


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    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] $


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    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.

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  3. #3
    Member
    Joined
    Dec 2008
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    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

    2 sina cosb answer

    2 sin (1/2)(x+y) + 2sin (1/2)(x-y) substituted........



    The product identities look tougher though because it looks like you have to memorize which identities to add....
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