Question 69630
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How would I express as a sum or difference

{{{2sin(6u)cos(4u)}}}

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You have to know two formulas:

{{{sin(A+B)=sin(A)cos(B)+cos(A)sin(B)}}}
{{{sin(A-B)=sin(A)cos(B)-cos(A)sin(B)}}}

Substitute (6u) for A and (4u) for B in both:

{{{sin(6u+4u)=sin(6u)cos(4u)+cos(6u)sin(4u)}}}
{{{sin(6u-4u)=sin(6u)cos(4u)-cos(6u)sin(4u)}}}

Combine terms in the parentheses on the left:

{{{sin(10u)=sin(6u)cos(4u)+cos(6u)sin(4u)}}}
 {{{sin(2u)=sin(6u)cos(4u)-cos(6u)sin(4u)}}}

Now add these two equations term by term

       {{{sin(10u)=sin(6u)cos(4u)+cos(6u)sin(4u)}}}
        {{{sin(2u)=sin(6u)cos(4u)-cos(6u)sin(4u)}}}
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{{{sin(10u)+sin(2u)=2sin(6u)cos(4u)}}}

That's the answer {{{sin(10u)+sin(2u)}}}

Edwin</pre>