Question 1003730
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{{{cos(3theta)sin(theta)}}}

This is the same as

{{{sin(theta)cos(3theta)}}}

We know that these identities hold:

{{{sin(theta+3theta)=sin(theta)cos(3theta)+cos(theta)sin(3theta)}}}

{{{sin(theta-3theta)=sin(theta)cos(3theta)-cos(theta)sin(3theta)}}}

We add those term by term, the last terms cancel and we get:

{{{sin(theta+3theta)+sin(theta-3theta)=2sin(theta)cos(3theta)}}}

{{{sin(4theta)+sin(-theta)=2sin(theta)cos(3theta)}}}

{{{sin(4theta)-sin(theta)=2sin(theta)cos(3theta)}}}

{{{expr(1/2)(sin(4theta)-sin(theta)^"")=sin(theta)cos(3theta)}}}

Or, distributing and rearranging:

{{{cos(3theta)sin(theta)=expr(1/2)sin(4theta)-expr(1/2)sin(theta)}}}

Edwin</pre>