Question 883546
{{{(sin(x)sin(2x)+sin(3x)sin(6x))/(sin(x)cos(2x)+sin(3x)cos(6x))=tan(5x)}}}

Let's break it down first by applying our product-to-sum formulae!
° {{{(sin(x)sin(2x))}}}={{{(cos(x-2x)-cos(x+2x))/2}}}={{{(cos(-x)-cos(3x))/2}}}
° {{{(sin(3x)sin(6x))}}}={{{(cos(3x-6x)-cos(3x+6x))/2}}}={{{(cos(-3x)-cos(9x))/2}}}
°{{{(sin(x)cos(2x))}}}={{{(sin(x+2x)+sin(x-2x))/2}}}={{{(sin(3x)+sin(-x))/2}}}
°{{{(sin(3x)cos(6x))}}}={{{(sin(3x+6x)+sin(3x-6x))/2}}}={{{(sin(9x)+sin(-3x))/2}}}

Remember:
* cos(-x)=cos(x)
*sin(-x)=-sin(x)

SO:

°{{{(cos(-x)-cos(3x))/2}}}={{{(cos(x)-cos(3x))/2}}}
°{{{(cos(-3x)-cos(9x))/2}}}={{{(cos(3x)-cos(9x))/2}}}
°{{{(sin(3x)+sin(-x))/2}}}={{{(sin(3x)-sin(x))/2}}}
°{{{(sin(9x)+sin(-3x))/2}}}={{{(sin(9x)-sin(3x))/2}}}

Now lets put it together!

{{{((cos(x)-cos(3x)+cos(3x)-cos(9x))/2)/(((sin(3x)-sin(x))+(sin(9x)-sin(3x)))/2)}}}

That looks kinda messy, so lets simplify that again and while we're at it, we'll get rid of one division by multiplying by the reciprocal!


{{{((cos(x)-cos(9x))/2)}}}*{{{(2/(sin(9x)-sin(x)))}}}

Cancel the 2's and we're left with:

{{{(cos(x)-cos(9x))/(sin(9x)-sin(x)))}}}

Now we're going to apply the sum-to-product formulae!

°{{{(cos(x)-cos(9x))}}}={{{(-2sin((x+9x)/2)*sin((x-9x)/2))}}}={{{(-2sin(5x)*sin(-4x))}}} Again, sin(-x)=-sin(x) so it's = to {{{(-2sin(5x)*(-sin(4x)))}}}
°{{{(sin(9x)-sin(x))}}}={{{(2cos((9x+x)/2)*sin((9x-x)/2))}}}={{{(2cos(5x)*sin(4x))}}}

Now we'll put the whole thing together:
 {{{(-2sin(5x)*(-sin(4x)))/(2cos(5x)*sin(4x))}}}={{{tan(5x)}}}
which simplifies to:

 {{{(sin(5x))/(cos(5x))}}}={{{tan(5x)}}}
and at last we have {{{tan(5x)=tan(5x)}}}