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January 31st, 2017, 12:16 PM   #1
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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.
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January 31st, 2017, 01:27 PM   #2
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$$2\cos(3x)\sin(x)=\sin(3x+x)-\sin(3x-x)=\sin(4x)-\sin(2x)$$
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January 31st, 2017, 02:52 PM   #3
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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.
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January 31st, 2017, 04:26 PM   #4
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The problem you posted is a specific case of a well known identity.
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January 31st, 2017, 07:02 PM   #5
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Quote:
Originally Posted by davedave View Post
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.
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January 31st, 2017, 11:57 PM   #6
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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?
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February 1st, 2017, 02:08 AM   #7
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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.
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February 2nd, 2017, 10:27 PM   #8
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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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