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How do you turn $14\sin(x)\cos(2x)$ into $7(\sin(3x)-\sin(x))$

how would one do this? How to do this?

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I get now use trig identities sin(3x)=sin(2x+x)=sin(2x)cos(x)+cos(2x)sin(x) and sin(x)=sin(2x-x)=sin(2x)cos(x)-cos(2x)sin(x) So using these two identities solve for cos(2x)sin(x) and add the equations

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    The two functions as written are not equal. – 2017-02-14
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    Try this: $\sin 3x = \sin (2x+x), \sin x = \sin (2x - x)$ Now apply the identities above. – 2017-02-14
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    http://mathworld.wolfram.com/WernerFormulas.html – 2017-02-14
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    do I replace sinxcos2x with sin3x-sin(2x)cosx? – 2017-02-14
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    @labbhattacharjee: Yep, that what the OP needs. By the way, in typical Precalculus and Calculus textbooks, these are usually called "Product-to-Sum Identities". – 2017-02-14

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