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 January 31st, 2017, 11:16 AM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 simplify without double angle identities I am going to write down a math problem below. (ex) Without using double angle identities, simplify 2cos(3x)*sinx my attempt: 2cos(3x)*sinx =2cos(x+2x)*sinx =2sinx*cosx*cos(2x)-2sinx*sinx*sin(2x) =2sinx*cosx*cos(2x)-2(sinx)^2*sin(2x) =2sinx*cosx*(cosx*cosx-sinx*sinx)-2sin(2x)+2(cosx)^2*sin(2x) =2sinx*(cosx)^3-2(sinx)^3*cosx-2sin(2x)+2(cosx)^2*sin(2x) My strategy makes the original expression very messy. I am really stuck. The answer is sin(4x)-sin(2x). Can someone explain how to do the problem without double angle identities? Thanks a lot.
 January 31st, 2017, 12:27 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $$2\cos(3x)\sin(x)=\sin(3x+x)-\sin(3x-x)=\sin(4x)-\sin(2x)$$ Thanks from davedave, agentredlum and topsquark
 January 31st, 2017, 01:52 PM #3 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 Thanks. greg1313. There is something I don't get. What made you use sin(3x+x) - sin(3x-x)? This is not obvious to me.
 January 31st, 2017, 03:26 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond The problem you posted is a specific case of a well known identity.
January 31st, 2017, 06:02 PM   #5
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Quote:
 Originally Posted by davedave Thanks. greg1313. There is something I don't get. What made you use sin(3x+x) - sin(3x-x)? This is not obvious to me.
It's because Greg knew that the sine of a compound angle gives products of sines and cosines of those angles.

 January 31st, 2017, 10:57 PM #6 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 Because I didn't know anything about this well-known identity, I had a hard time with the problem. How would you approach it if you didn't know that identity?
 February 1st, 2017, 01:08 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 Alternatively, use $\cos(\theta) = (e^{i\theta} + e^{-i\theta})/2$, $\sin(\theta) = (e^{i\theta} - e^{-i\theta})/(2i)$, and the law of exponents. The task then reduces to easy algebra. Thanks from davedave
 February 2nd, 2017, 09:27 PM #8 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 This problem "looks" easy, but it requires you to use advanced ideas to solve it like the polar form of sine and cosine. I wish there was a much easier way to do it. Thanks, everyone.

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