Question 920256

Let {{{Q = x+y}}}. You don't have to use Q. Just any other variable not being used.


{{{sin(x+y+z)}}} will then turn into {{{sin(Q+z)}}} after you replace "x+y" with "Q" (the substitution is valid based on the definition we made above).


Expand out {{{sin(Q+z)}}}:


{{{sin(Q+z) = sin(Q)cos(z)+cos(Q)sin(z)}}}


Now we have {{{sin(Q)cos(z)+cos(Q)sin(z)}}}. Next, replace every Q with "x+y" to get 


{{{sin(Q)cos(z)+cos(Q)sin(z)}}}


{{{sin(x+y)cos(z)+cos(x+y)sin(z)}}} replace every Q with "x+y"


The next step is to expand {{{sin(x+y)}}} and {{{cos(x+y)}}}


{{{sin(x+y) = sin(x)cos(y)+cos(x)sin(y)}}}
{{{cos(x+y) = cos(x)cos(y)-sin(x)sin(y)}}}


which you will use to replace {{{sin(x+y)}}} and {{{cos(x+y)}}} in {{{sin(x+y)cos(z)+cos(x+y)sin(z)}}}


{{{sin(x+y)cos(z)+cos(x+y)sin(z)}}}


{{{(sin(x)cos(y)+cos(x)sin(y))cos(z)+cos(x+y)sin(z)}}} Replace {{{sin(x+y)}}} with {{{sin(x)cos(y)+cos(x)sin(y)}}} 


{{{(sin(x)cos(y)+cos(x)sin(y))cos(z)+(cos(x)cos(y)-sin(x)sin(y))sin(z)}}} Replace {{{cos(x+y)}}} with {{{cos(x)cos(y)-sin(x)sin(y)}}}


{{{sin(x)cos(y)cos(z)+cos(x)sin(y)cos(z)+cos(x)cos(y)sin(z)-sin(x)sin(y)sin(z)}}} Distribute



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So {{{sin(x+y+z)}}} expands out into {{{sin(x)cos(y)cos(z)+cos(x)sin(y)cos(z)+cos(x)cos(y)sin(z)-sin(x)sin(y)sin(z)}}}



Let me know if you need more help or if you need me to explain a step in more detail.
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Thanks,


Jim